Supp Textbook Icfai

Operations Management Supplement to Textbook

Icfai Center for Management Research

Road # 3, Banjara Hills, Hyderabad 500 034

ii

Icfai, March 2007. All rights reserved.

No part of this publication may be reproduced, stored in a retrieval system, used in a spreadsheet, or transmitted in any form or by any means – electronic, mechanical, photocopying or otherwise – without prior permission in writing from Icfai.

Ref. No. SOM – 03 2K7 35 For any clarification regarding this book, the students may please write to Icfai giving the above reference number, and page number. While every possible care has been taken in preparing this book, Icfai welcomes suggestions from students for improvement in future editions.

iii

SUPPLEMENT TO OPERATIONS MANAGEMENT (Ref. No. of Supplement: SOM - 03 2K7 35)

NOTE:

• The 1st Edition of the Operations Management textbook (Ref. No.: OM - 052K3 35 or OM - 09 2K3 35) should be read along with this supplement (Ref. No.: SOM – 03 2K7 35).

• The 1st Edition of the textbook, along with this supplement, is equivalent to the 2nd Edition of the Operations Management textbook (Ref. No.: OM – 03 2K7 35).

• H1, H2, H3 refers to first level, second level and third level headings in the text book.

CHAPTER TITLE TEXT ADDED PROBLEMS ADDED

Chapter 1 Operations Management – An Overview None None

Chapter 2 Operations Strategy ! Financial and Economic

Analysis in Operations (H1) None

Chapter 3 Forecasting Demand ! Forecast Components (H1)

! Demand Forecasting

Process (H1)

! Time Series Methods (H2)

! Simple moving average (SMA) (H3)

! Weighted moving average (H3)

! Trend adjusted exponential smoothing

(double smoothing) (H3)

! Least square method (H3)

Chapter 4 Allocating Resources to Strategic

Alternatives

! Corrections in the 1st

Edition (Errata)

! Graphical method (H2)

Do

Not

Cop

y

iv


Chapter 5 Design of Production Processes None None

Chapter 6 Facility Location and Layout ! Steps in assembly line

balancing (H3)


Edition (Errata)

! Cost-Profit-Volume or Break-Even

Analysis (H2)

Chapter 7 Job Design (Chapter 10 in the 1st Edition) None None

Chapter 8 Work Measurement (Chapter 11 in 1st

Edition)

None None

Chapter 9 Aggregate Planning and Capacity

Planning

(Chapter 7 in the 1st Edition)

! Measuring capacity (H2)

! Determining capacity

requirements (H2)

! Economies of scale (H2)

! Varying the size of inventory (H3)

! Make-to-order items (H3)

Chapter 10 Fundamentals of Inventory Control


! Inventory Classification

Models (H1)

! Optimal Order Quantity (H2)

Chapter 11 Purchase Management (Chapter 9 in the

1st Edition)

None ! Make-or-Buy Decisions (H1)

Chapter 12 Materials Management (Chapter 13 in the

1st Edition)


Edition (Errata) None

Do

Not

Cop

y

v


Chapter 13 Materials Requirement Planning (Chapter

14 - Inventory Management: Dependent

Demand - in the 1st Edition)

None ! Bills of material (H3)

! Offsetting (H3)

Chapter 14 Operations Scheduling (Chapter 17 in the

1st Edition)

! Scheduling Consecutive

Days Off (H2)

! Scheduling Daily Work

Times (H2)

! Scheduling Hourly Work

Times (H2)

! Backward Scheduling (H2)

! Johnson’s rule for n-jobs and m-

machines (H2)

! Critical Ratio Method (H2)

Chapter 15 Enterprise Resource Planning None None

Chapter 16 Supply Chain Management None None

Chapter 17 Just-In-Time (JIT) Manufacturing System


None None

Chapter 18 Productivity and Quality Management

(Chapter 19: Quality Management in the

1st Edition)

! Productivity (H1)

! Control charts for variables

(H2)

! Control charts for attributes

(H2)

! TQM Principles (H2)

! Control charts for variables (H2)

! Control charts for attributes (H2)

Do

Not

Cop

y

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Chapter 19 Facilities and Maintenance Management


! Facilities Management (H1)

! Maintenance Planning (H1)

! Evaluation of Preventive Maintenance

Policies (H1)

Chapter 20 Project Management (Chapter 21 in the 1st

Edition)

None ! Computation of latest times: backward

pass (H3)

! PERT (H2)

Chapter 21 Trends in Operations Technology


None None

Chapter 22 Globalization and Operations

Management

None None

D

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ot C

opy

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Contents Supplement to Textbook

Operations Strategy 1-2

Forecasting Demand 3-16

Allocating Resources to Strategic Alternatives 17-20

Facility Location and Layout 21-27

Aggregate Planning and Capacity Planning 28-32

Fundamentals of Inventory Control 33-34

Purchase Management 35-36

Materials Management 37-38

Materials Requirement Planning 39-41

Operations Scheduling 42-51

Productivity and Quality Management 52-63

Facilities and Maintenance Management 64-74

Project Management 75-77

Operations Strategy • Operations Strategy as a Competitive Weapon • Elements of Operations Strategy • Developing an Operations Strategy • Financial and Economic Analysis in Operations (New)

FINANCIAL AND ECONOMIC ANALYSIS IN OPERATIONS

(Added after ‘Developing an Operations Strategy’ in page 16 of the first edition)

Operations managers have to take both financial and economic decisions. Economic and financial analysis is used to evaluate the costs of operation and profit potential of an investment. Operations managers should be familiar with conditions and factors affecting costs and the methods of measuring and controlling costs. Operations costs are divided into direct costs and indirect costs. Direct costs or prime costs are those cost components, which can be identified individually for each product or service produced, e.g. the cost of direct material, the cost of direct labor etc. Indirect costs or operations overhead are those which cannot be tied to specific product or service e.g. administrative costs, maintenance costs. In addition, an operations manager should be aware of the fixed and variable components of the costs. Fixed costs are those which do not change with a minor change in scale of production e.g. rent on premises, depreciation on equipment, insurance etc. Variable costs denote direct costs like wages of workers, and direct material cost which vary with the change in the scale of production. Operations managers have many methods at their disposal to evaluate the cost effectiveness of an investment. Two of the most commonly used techniques are the payback method and net present value (NPV) method.

Payback Method One of the basic methods used to compare investment alternatives is by calculating the payback period for each investment alternative. The payback period is the time taken (usually in years) to recover the initial investment. The payback takes into consideration the initial investment and the resulting annual cash flow. Mathematically the payback period is denoted by:

investment from income annualNet investmentNet periodPayback =

Where,

Payback period = Time taken to recover initial investment (usually in years)

Net Investment = purchase and installation costs minus its anticipated future salvage value

Net Annual income = Anticipated annual revenue minus expenses

In spite of its popularity, the payback method has a few inherent drawbacks - it ignores cash flow beyond the payback period, and does not take into account the time value of money. These deficiencies are overcome in the net present value method.

Operations Management

2

For example, let us calculate the payback period for a project with an initial investment of Rs 10 lakh that is expected to generate an income of Rs 2 lakh per annum. The payback period or the time taken to recover the initial investment can be calculated using the formula given below

income) annual (expectedlakh 2)Investment(Net lakh 10

Payback =

When a firm needs to decide on a particular investment from various alternatives, the payback period for each investment alternative is calculated. The alternative with the shortest payback period is preferred.

Net Present Value (NPV) Method The NPV method is used to calculate present value of future returns, discounted at the marginal cost of capital, minus the present value of the cost of the investment. The net present value method takes into account the time value of money. It is employed for ranking and comparing the profitability of project alternatives. NPV for a project can be determined by using the equation: Where,

CFi = the cash flow at time i r = discount rate t = time horizon I = Initial investment If the net present value of an investment is greater then one then the project is acceptable. If the net present value is less than zero, the project is rejected. The greater the NPV value of the project the better is its profitability. Incase, where multiple projects are compared the project alternative with the largest NPV is selected. Suppose the net present value of a project is calculated as Rs 1lakh, this implies that undertaking the project is expected to increase the value of the firm by Rs 1 lakh. NPV is a useful method for comparing investment alternatives, with comparable initial investments.

Iir)(1

iCFNPV

n

1i

−−

∑=

=

Forecasting Demand • Forecasting in Operations

• Forecast Components (New)

• Demand Forecasting Process (New)

• Forecasting Methods (Modified)

• Selecting a Forecasting Method

• Measures of Forecasting Accuracy

• Monitoring and Controlling Forecasts [

FORECAST COMPONENTS

(After ‘Forecasting in Operations’ in page 20 of the first edition) For forecasting purposes, firms need to take into consideration various factors, or components. For any product, there are six major forecast components: base demand, seasonal factors, trends, cyclical factors, promotions, and the irregular component.

All the components need not be included in each forecast – however, all of them have to be monitored and considered. Let us examine each of the six components and their characteristics. Base demand: Base demand is the average of sales over a given time period. This figure can be taken as the right forecast if the products' demand is not impacted by the seasonal, trend, cyclic and promotional factors.

Seasonal component: This refers to the repeated pattern of increase and decrease in demand over a period of time. For example, the demand for cool drinks and ice creams peaks during summer and remains low during the rest of the year. In this case, the demand pattern shows high seasonal variability during summer and low seasonal variability during rest of the year.

Trend component: The trend component refers to the long term pattern of movement of demand over a period of time. The trend may be positive, negative or neutral. In any given time period, a positive trend implies that the demand is increasing, while a negative trend indicates that the demand is decreasing. A neutral trend implies that changes in the demand are negligible over a period of time. For example, the motor bikes sales in India have been showing a positive trend since the late 1990's due to changes in customer preferences. On the other hand, the trend in scooter sales in India has changed from positive to negative during the same time period.

Cyclic component: Cyclic component refers to changes in the demand patterns, which exist for more than one year. These changes could either show an upward or a downward movement. A good example of the cyclic component would be the business cycle, which changes every few years (from recession to boom and vice-versa). The demand for luxury products may be linked with the business cycle, since sales usually increase during the boom phase and slow down during recessions.

Promotional component: The promotional component is one of the key factors that impact demand, especially for FMCG and consumer appliances companies, whose sales are more dependent on promotions. It refers to the changes in demand that occur due to the promotional campaigns undertaken by the firm. Promotions can be for the


4

end customers and/or for the distributors and retailers. Their impact on the forecast also depends on their regularity. A regular promotion is run at the same time every year. Discount sale by a refrigerator manufacturer during the lean season can be an example here. The regular promotion component can be termed as seasonal component from the forecasting perspective. Irregular promotions are those promotions which are run as per the market conditions and do not have a specific time frame. Such promotions need to be monitored carefully and considered separately. Unlike other forecasting components, the promotional component is controllable by the firm. Irregular component: The irregular component refers to all those variations in demand that cannot be attributed to any of the above five factors. This factor is difficult to predict because of its random nature. Any forecasting exercise involves efforts to reduce the impact of this component by accurately monitoring and predicting the other components.

DEMAND FORECASTING PROCESS

(After ‘Forecast Components’ given above)

Understand the Objective of Forecasting It is extremely important for a firm to be clear about the objectives of the forecasting process. For every firm, the motive behind carrying out a forecasting exercise is to implement decisions which are based on the forecast. Thus, first the firm should identify the decisions that need to be implemented. It should also appraise all the supply chain members concerned about these decisions. For example, if HP plans to run a discount sale, then the dealers and logistics partners need to be aware of this plan, so that they can meet the demand effectively. All the participants should arrive

Exhibit 3.1 Collaborative Forecasting

To fully realize the benefits of forecasting, the different departments of an organization should integrate their forecasting objectives. Separate forecasting methods, technologies and objectives for each functional department (like finance, marketing, sales and production) can lead to a waste of resources, missed opportunities, etc. due to multiple forecasts. An organization can improve its overall forecast by adopting a collaborative approach. Collaborative forecasting involves the synchronization of the different forecasting objectives and systems employed by various functional entities. An organizational forecasting system should contain a central database that gathers information from enterprise systems like ERP, SCM, and CRM. These systems do not have an in-built functionality that caters to the requirements of collaborative forecasting. In order to obtain a single forecast that meets the needs of all the departments, information from these systems should be utilized to generate a forecast that meets the overall requirements of the organization and its different functions. In order to ensure that the single forecast takes into account the specific requirements of different departments or functions, a consensus methodology is used. According to the Gartner Group, “Enterprises that collaboratively integrate disparate forecasting systems will improve revenue predictability by 10% to 25% and decrease inventory carrying costs by more than 30% over a three year period.”

Adapted from Michael J Hennel, “Forecasting Demand Begins with Integration” B to B, 11/11/2002, Vol. 87 Issue 11, p9, 1p.

Forecasting Demand

5

at a consensus about the forecast for the promotion period and decide upon a common action plan. If the members are not able to plan jointly, then mismatches between the demand and supply would arise at various levels of the supply chain. This could also result in logistics problems since transporters and warehouse operators would not have timely access to the promotion plan, resulting in a shortage of transportation vehicles and/or warehouse space to meet the increasing customer demand.

Integrate Demand Planning and Forecasting

The forecast is the basis for most of the planning activities such as capacity planning, production planning, promotion planning and various other functions like procurement and distribution. Hence, the firm has to integrate all these aspects right from the initial stages of working out the forecast. The above can be achieved both at the information systems level as well as at the human resource management level. To facilitate easy integration of various functional areas, the firm can form cross-functional teams representing all the departments. Many firms adopt collaborative forecasting so as to integrate the planning and forecasting of different functions like marketing, production, etc. This is done through integrating the objectives and systems employed by these functions. Refer Exhibit 3.1 for further details on collaborative forecasting.

Identify Major Factors that Influence the Demand Forecast

The next step in the forecasting process is to determine the key factors that influence the demand forecast. These factors have already been discussed in the forecasting considerations section. The key task for the firm here would be to identify the major components that impact the forecast. For instance, if the firm is dealing with seasonal products, then seasonal factors would have a larger influence on the demand. Similarly, if the firm is dealing in FMCG products, then the promotional factor would play a greater role in influencing the demand.

Understand and Identify Customer Segments

The next step is identifying the customer segments. Under this, firms identify customers with similar service requirements, demand pattern, seasonality etc., and cluster them into segments. For each such customer segment, the most appropriate forecasting method should be selected.

Determine the Appropriate Forecasting Technique

Selecting an appropriate forecasting technique is a crucial step in the demand forecasting process, where the firm has to analyze various issues. These may include the stage of product life cycle, the geographical region and the customer group etc. If the product is in the early stages of its life cycle (i.e., when the product has been launched recently), then it would have little or no sales history. In such cases, judgmental techniques would be more appropriate. But if the product is in the maturity stage, time series techniques can be used since a lot of data would be easily available. If detailed sales related data is available for a product, then causal techniques can be a right option.


6

FORECASTING METHODS

Time-Series Methods

(Added under ‘Time Series Methods’ in page 23 of the first edition) Time-series forecasting methods assume that past data is a good indicator of the future. Instances where this assumption is not true are rare and not significant enough. Some relevant data can always be found. Hence, many operations managers use a time series model to forecast the demand for their goods or services.

Time series is a widely used forecasting method, largely because it is simple and inexpensive to use. Forecasts obtained from this method can be a good basis for further analysis. The term 'time series' refers to a collection of well-defined data, which is obtained through repeated measurement over a period of time. Weekly or monthly sales at each retail outlet of a company would qualify under this definition. The collected data is said to be well defined because it is collected at regular time intervals. Thus, data collected randomly or irregularly cannot be called as a time series.

Time series analysis can be categorized into two broad categories, based on the complexity involved: static and adaptive.

Static forecasting methods

These methods are also known as basic time series forecasting techniques. Static forecasting methods assume that the estimates of trend and seasonal components do not vary from year to year. In this method, estimates of the trend and seasonal components are determined based on historical data, which is projected to obtain future demand data. Static forecasting methods follow the procedure given below to obtain forecasts: • Deseasonalizing or decomposing the time series

• Identifying the trend and seasonal components

• Making the forecasts based on trend and seasonal components Let us explore the static forecasting technique with the help of an example. Consider a consumer appliances company, ABC Limited that is planning to forecast future sales based on the analysis of past demand data. The firm has collected sales information for the past five years on a quarterly basis, as given in Table 3.2.

Table 3.2: Quarterly Sales of ABC Limited in ‘000 units

Year I II III IV

1998 40 25 15 30

1999 38 26 13 28

2000 43 22 18 33

2001 45 29 14 30

2002 41 26 18 29

As mentioned earlier, in order to forecast the future demand, the following steps need to be performed: deseaonalizing the time series, estimating the trend and seasonal components and making the forecast.

Forecasting Demand

7

Deseasonalizing the time series

Deseasonlizing or decomposing the time series refers to identifying the seasonal variations in the time series and removing those effects. For this, first calculate the seasonal index. Later, using the seasonal indices, the effects of seasonality are removed from the time series. Calculating seasonal index: Steps involved in calculating the seasonal index are as follows:

Step1: First the four quarter moving total for the time series is calculated. The sum of four quarters of the year 1998 i.e. 40+25+15+30 = 110 is calculated and placed at the mid data point of four values of the year 1998, between the II quarter and III quarter (See the column four of Table 3.3). Then the moving total is calculated for the next set of four values by dropping the first value i.e. the value of quarter-I of 1998 and adding the value of quarter-I of 1999. So, the sum of the four values i.e. 25+15+30+38 is 108 and placed between the values of III and IV quarters of 1998. The next sets of four values are III and IV quarters of 1998 and I and II quarters of 1999. The total i.e., 25 + 15 + 30 + 38 = 108, is placed between the IV quarter of 1998 and I quarter of 1999. This process of moving the four quarters and calculating the total is continued till the last value of the time series is included in the calculation. The values of four quarters of 2002 are the last set of values. The total of these four quarters is placed between the II and III quarters of 2002.The resultant values are shown in column four of Table 3.3.

Table 3.3: Calculation of 4-Quarter Center Moving Averages Year (1) Quarter

(2) Actual

Sales (3) 4-quarter

moving totals

(4)

4-quarter moving

averages (5)= (4)/4

4-quarter centered moving

averages (6)

Percentage of actual to

moving average

values (7)

1998 I 40

II 25

110 27.5

III 15 27.25 55.05%

108 27

IV 30 27.125 110.60%

1999 109 27.25

I 38 27 140.74%

107 26.75

II 26 26.5 98.11%

105 26.25

III 13 26.875 48.37%

110 27.5

IV 28 27 103.70%

2000 106 26.5

I 43 27.125 158.53%


8

111 27.75

II 22 28.375 77.53%

116 29

III 18 29.25 61.54%

118 29.5

IV 33 30.375 108.64%

2001 125 31.25

I 45 30.75 146.34%

121 30.25

II 29 29.875 97.07%

118 29.5

III 14 29 48.28%

114 28.5

IV 30 28.125 106.67%

2002 111 27.75

I 41 28.25 145.13%

115 28.75

II 26 28.625 90.83%

114 28.5

III 18

IV 29

Step 2: Then the 4-quarter moving average is calculated. This is obtained by dividing the values of moving totals (obtained in column four of Table 3.3) by four. So moving average of first set of values is 110/4 = 27.5. Similarly, moving averages are calculated for the rest of the values in column four of Table 3.3. These values are shown in column five of Table 3.3.

Step 3: Then the moving averages are centered. The moving averages in column five fall halfway between the quarters of each year. Average of two four-quarter moving averages falling just above and below the quarter is calculated to center out the averages. For quarter-III of 1998, the resulting four-quarter centered moving average is 27.25. This was obtained by taking average of two moving averages i.e. (27.5 + 27)/2 = 27.25. Similarly, the other values of column five of Table 3.3 are centered out which are shown in column six of Table 3.3. (Step 3 is not required if the number of values is odd as the values will be centered out in the step 2 itself.)

Step 4: For each quarter which is having a 4-quarter centered out moving average, the percentage of actual value to the moving average is calculated. The step is performed to regain the seasonal component of the quarters. The percentage is obtained by dividing the actual quarter value with each associated 4-quarter centered moving average value. Then the resulting value is multiplied by 100.

Therefore, for quarter III of 1998, the value is (15/27.25) × 100 = 55.05%. Similarly, other values are calculated and shown in column seven of Table 3.3.

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Step 5: The values obtained in column seven are arranged by quarter (as shown in Table 3.4), and the modified mean is calculated. The modified mean is obtained by removing the highest and lowest values for each quarter and taking average of the remaining values.

For example, for quarter I, the highest and lowest values are 158.53 and 140.74 respectively. By canceling these and totaling the remaining values we get the modified sum as 291.47. Then, the modified mean of quarter I is 291.47/2 = 145.74

Similarly for quarter II, the highest and lowest values are 98.11 and 77.53 respectively. By canceling and totaling the remaining values we get the modified sum as 187.9. Thus modified mean of quarter II is 187.9/2 = 93.95

Table 3.4: Calculating Modified Mean Year Quarter I Quarter II Quarter III Quarter IV 1998 -- -- 55.05 110.60 1999 140.74 98.11 48.37 103.70 2000 158.53 77.53 61.54 108.64 2001 146.34 97.07 48.28 106.67 2002 145.13 90.83 -- --

Modified sum 291.47 187.90 103.42 215.31

Modified mean 145.74 93.95 51.71 107.65

By removing the highest and lowest values for each quarter, the extreme irregular variations are removed. By averaging the rest of the values, the time series is further smoothed. Thus the modified mean represents the seasonality component. Step 6: In the final step, the modified mean is further adjusted to obtain the seasonal index. From Table 3.4, the total of 4 indices = 145.74 + 93.95 + 51.71 + 107.65 = 399. But 100 is the base for the index. Hence the total of four quarterly indices should be 400, and mean of the indices needs to be 100. In order to rectify the error, the seasonal indices are adjusted by multiplying each index by an adjusting constant. The adjusting constant is obtained by dividing the desired index total i.e. 400 by actual total of 399. The value obtained is 400/399 = 1.0025. The process of adjusting the modified mean and obtaining seasonal index is shown in Table 3.5.

Table 3.5: The Calculation of Seasonal Indices

Quarter (1)

Index (2)

Adjusting constant (3)

Seasonal Index (4) = (2) x (3)

I 145.74 1.0025 146.10

II 93.95 1.0025 94.18

III 51.71 1.0025 51.83

IV 107.65 1.0025 107.92

Seasonal indices for different quarters are shown in the column four of Table 3.5. The next step is calculating the deseasonalized time series values. To obtain deseasonalized time series values, the actual sales values are divided by the resultant value achieved by dividing corresponding seasonal indices with 100. Deseasonalized sales are calculated for various quarters in Table 3.6


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Table 3.6: Calculating Deseasonalized Sales Values Year (1)

Quarter (2)

Actual Sales (3)

Seasonal Index/100 (4)

Deseasonalized Sales (5) = (3)/(4)

1998 I 40 1.461 27.38

II 25 0.942 26.54

III 15 0.518 28.96

IV 30 1.079 27.80

1999 I 38 1.461 26.01

II 26 0.942 27.61

III 13 0.518 25.10

IV 28 1.079 25.95

2000 I 43 1.461 29.43

II 22 0.942 23.36

III 18 0.518 34.76

IV 33 1.079 30.58

Year (1)

Quarter (2)

Actual Sales (3)

Seasonal Index/100 (4)

Deseasonalized Sales (5) = (3)/(4)

2001 I 45 1.461 30.80 II 29 0.942 30.79 III 14 0.518 27.03 IV 30 1.079 27.80

2002 I 41 1.461 28.06 II 26 0.942 27.61 III 18 0.518 34.76 IV 29 1.079 26.87

Developing trend component After depersonalizing the time series, a trend component of the time series is calculated. This can be done using the least squares method. The time variable (years 1999, 2000, etc. in this case) is converted into a form which would simplify the calculation. This conversion of the time variable is accomplished using a technique called coding. In this technique, mean of all the sample times is calculated, and then the resultant value is subtracted from each of the sample times.

Table 3.7: Calculations for Developing the Trend Equation Year (1)

Quarter (2)

y Deseasonalized Sales (Column 5 of Table 3.6)

(3)

Translating or Coding the

Time Variable (4)

x Coded Time

Variable (5) = (4) × 2

xy (6) = (5) × (3)

x2 (7) = (5)2

1998 I 27.38 -9 ½ -19 -520.22 361 II 26.54 -8 ½ -17 -451.18 289

Forecasting Demand

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III 28.96 -7 ½ -15 -434.4 225 IV 27.80 -6 ½ -13 -361.4 169

1999 I 26.01 -5 ½ -11 -286.11 121 II 27.61 -4 ½ -9 -248.49 81 III 25.10 -3 ½ -7 -175.7 49 IV 25.95 -2 ½ -5 -129.75 25

2000 I 29.43 -1 ½ -3 -88.29 9 II 23.36 - ½ -1 -23.36 1

Mean 0

III 34.76 ½ 1 34.76 1

IV 30.58 1 ½ 3 91.74 9

2001 I 30.80 2 ½ 5 154 25

II 30.79 3 ½ 7 215.53 49

III 27.03 4 ½ 9 243.27 81

IV 27.80 5 ½ 11 305.8 121

2002 I 28.06 6 ½ 13 364.78 169

II 27.61 7 ½ 15 414.15 225

III 34.76 8 ½ 17 590.92 289

IV 26.87 9 ½ 19 510.53 361

∑ y = 567.2 ∑ xy =

206.58 ∑ 2x =

2,660

The idea behind this exercise is to eliminate the need for squaring large numbers like year 2000, 2001. By setting the mean as zero, the calculations are simplified to a great extent. Suppose we are dealing with three time points - year 2000, 2001 and 2002. We assign the codes -1, 0, 1 to the respective years, where zero represents the mean (year 2001), -1 represents the last or previous year (2000) and 1 represents the next year (2002). This translation process will differ depending upon the number of time periods. If the problem consists of an odd number of time periods, the mean can be easily found out. For instance, if there are 7 time points, then the mean would be the middle of the time points i.e. 4. However, if the time points are even in number, then the mean would contain a ‘½ fraction,’ which if subtracted from each time point, would make the calculations cumbersome. To overcome this problem, each time variable is multiplied with 2. The coded variable is denoted by x. In this problem, the time points are 20, the mean is 9½ which falls between the II quarter and III quarter of 2001. The first value i.e., Quarter I of 1998 is assigned as -9½ and the last value i.e. quarter IV of 2002 is assigned as 9½. The translating time variable is shown in column four of Table 3.7, and the coded time points, which are obtained by multiplying the translating variables with two, are shown in column five of Table 3.7. Now using the least squares method, trend component is developed. The trend component is obtained using the equation:


12

bx a +=∧y -----------Equ 3.1

Where ∧

y = estimated value of the dependent variable

x = independent variable (time in trend analysis)

a = y-intercept (the value of y when x = 0) = y = y/n

b = slope of the trend line = ∑∑

2xxy

By substituting the values of xy, x2, y and n, the values of a and b are 28.36 and 0.08 respectively.

Therefore, ∧

y = 28.36 + 0.08x ------------------Equ 3.2

Now, various values of x can by substituted in Equ 3.1 and the estimated values of ∧

y for each value of x. The calculation is shown in Table 3.8.

Making the forecast

The final step in the static forecasting method is to forecast the demand for future periods. The following steps are followed in forecasting the demand.

Step 1: Calculate the trend level for the periods for which forecast needs to be made using the trend equation.

Step 2: Next step is to multiply the trend level from Step 1 with the period seasonal index to include seasonal effects.

Step 3: Then multiply the result of Step 2 by the projected cyclic index to include cyclic effects and get the final forecast result.

Suppose the firm wants to forecast for all the four quarters of year 2003.

First, all the four quarters of year 2003 need to be coded. The coded value after multiplying by two for the last quarter (IV) of 2002 is 19. Looking at the trend in Table 3.7, it can be easily estimated that the coded time variable for the four quarters of 2003 will be 21, 23, 25 and 27 respectively.

Substituting these values in Equ 3.2, we get ∧

y for I, II, III and IV quarter of 2003 is 30.04, 30.2, 30.36 and 30.52 respectively.

Next step is to calculate the seasonalized estimates by multiplying the deseasonalized sales estimates with the respective seasonal indices and expressing them as a fraction of 100. Seasonal indices for the four quarters are shown in Table 3.5.

Seasonalized estimates for quarter I of 2003: 30.04 × (146.1/100) = 43.89 Seasonalized estimates for quarter II of 2003: 30.2 × (94.18/100) = 28.44 Seasonalized estimates for quarter III of 2003: 30.36 × (51.83/100) = 15.72 Seasonalized estimates for quarter IV of 2003: 30.52 × (107.92/100) = 32.94

Forecasting Demand

13

Table 3.8: Trend Component Calculation

x (From column 5 of Table 3.7) ∧

y = a +bx = 28.36 + 0.08x

-19 28.36+0.08(-19) =26.84

-17 28.36+0.08(-17) = 27

-15 28.36+0.08 (-15) = 27.16

-13 28.36+0.08 (-13) = 27.32

-11 28.36+0.08 (-11) = 27.48

-9 28.36+0.08 (-9) = 27.64

-7 28.36+0.08 (-7) = 27.8

-5 28.36+0.08 (-5) = 27.96

-3 28.36+0.08 (-3) = 28.12

-1 28.36+0.08 (-1) = 28.28

1 28.36+0.08 (1) = 28.44

3 28.36+0.08 (3) = 28.6

5 28.36+0.08(5) = 28.76

7 28.36+0.08(7) = 28.92

9 28.36+0.08(9) = 29.08

11 28.36+0.08(11) = 29.24

13 28.36+0.08(13) = 29.4

15 28.36+0.08(15) = 29.56

17 28.36+0.08(17) = 29.72

19 28.36+0.08(19) = 29.88

Adaptive forecasting Adaptive forecasting is an advanced form of time series analysis, where the trend and seasonal components are adjusted after each demand observation. There are three widely used adaptive forecasting methods. They are simple moving average, weighted moving average, and exponential smoothing.

FORECASTING METHODS

Simple moving average (SMA) (Added under Simple Moving Average in page 24 of the first edition)

Problem A company’s demand for liquid crystal displays (LCDs) for the 12 months from January 2006 to December 2006 has been given in the following table. Calculate the demand forecast for LCDs for January 2007 using a six-month simple moving average forecasting technique.


14

Months Jan Feb Mar Apr May Jun July Aug Sep Oct Nov Dec

Demand (in lakhs) 53 53 56 57 58 56 58 58 60 61 59 60

Solution Using a six-month simple moving average technique, the forecast for the seventh month can be calculated as the average of the first six months. Thus, the forecast for July 2006 is the average demand of LCDs during January, February, March, April, May, and June. Similarly, the forecast for January 2007 would be the average of the demand during July, August, September, October, November, and December 2006.

Forecast for January 2007 = (58 + 58 + 60 + 61 + 59 + 60) / 6 = 356/6 = 59.3 lakhs

Thus, by using moving averages for six months, we obtain the demand for the month of January 2007 as 59.3 lakhs.

FORECASTING METHODS

Weighted moving average

(Added under ‘Weighted Moving Average’ in page 25 of the first edition)

Problem The demand for tuna fish for the months of April, May, June, July, August, and September are 150, 155, 158, 162, 164, and 163 metric tons respectively. Calculate the demand for October using the three-month weighted moving average technique. Assign the following weights to the data: 50% weight for September, 30% for August, and 20% for July.

Solution Using the three-month weighted moving average, the demand for October can be calculated as follows:

Forecast for October (WMA3) = (demand in July x 0.20) + (demand in August x 0.30) + (demand in September x 0.50)

= (162 x 0.20) + (164 x 0.30) + (163 x 0.50)

= 32.4 + 49.2 + 81.5

= 163.1 metric tons

Thus, the demand for tuna fish for October arrived at by using the three-month weighted moving average is 163.1 metric tons.

FORECASTING METHODS

Trend adjusted exponential smoothing (double smoothing) (Added under ‘Trend adjusted exponential smoothing (double smoothing)’ in Pg 27 of the first edition) Problem Hansa Cycles Ltd. manufactures cycles of different models for children, men, and women. The sales for its high-end sports model for children averaged 2500 units for the last 3 years. The average increase in sales was 100 units per year for the model. Last year, 2650 units were sold. Forecast the sales for this year, using the trend

Forecasting Demand

15

adjusted exponential smoothing method. Take smoothing constants as α = 0.3 and β = 0.25.

Solution From the information given, A0 = 2500 and T0 = 100

D1 = 2650

To forecast sales for this year, we will use the equations

At = αDt + (1−α)(At-1+Tt-1)

Tt = β(At−At-1) + (1−β)(Tt-1)

Ft+1 = At + Tt

A3 = 0.3(2650) + (1 − 0.3)(2500 + 100) = 795 + 0.7(2600) 795 + 1820 = 2615

T3 = 0.25(2615 − 2500) + (1 - 0.25)(100) = 28.75 + 75 = 103.75

F4 = 2615 + 103.75 = 2718.75

Hence, the demand for the sport cycle model for children in the current year would be 2718.75 (2719) units.

FORECASTING METHODS

Causal Quantitative Models Least square method (Added under ‘Least square method’ in page 29 of the first edition) Problem A firm would like to forecast the cost of raw materials needed to manufacture 50,000 units of product A. Help the firm calculate the cost of raw materials using regression analysis. The production for the last three years and the corresponding costs of raw materials are given in the following table. Here, sample size ‘n’ = 3.

Product A (in `000) 40 44 48 Cost of Raw Materials (in Rs.`000) 12 13 16

Solution

Let us first calculate the values of ‘a’ and ‘b’ using the least square method. Consider production units as X and cost of raw materials as Y.

X (in `000) Y (in Rs.`000) XY X2 Y2

40 12 480 1600 144

44 13 572 1936 169

48 16 768 2304 256

132 41 1820 5840 569

X = n

X∑ = 132 / 3 = 44 (n =3)


16

Y = n

Y∑ = 41 / 3 = 13.67

∑ ∑

∑∑−∑=

2X)( - )2X

Y)(X)(XY)n(b

(n

Now, substituting the values of ∑XY, ∑X2, and ∑Y2 from the table, we get b = 3(1820) – [(132)(41)] / (3(5840) – (132)(132)) b = (5460 – 5412) / (17520 – 17424) b = 48/96 b = 0.5 Next, we calculate the value of constant ‘a’ using the equation:

a = XbY − = 13.67 – 0.5(44) a = -8.33 Substituting the values of ‘a’ and ‘b’ in the straight line equation, we can obtain the value of the dependent variable on the basis of the independent variable. Then, we can find the costs of raw materials required (dependent variable) for producing 50,000 units of product A. Y = a + bX Y = -8.33 + 0.5× 50 Y = 16.67 Thus, for producing 50,000 units, the cost of raw materials would be Rs.16,670.

Allocating Resources to Strategic Alternatives

• Allocation Decisions in Operations Strategy • Linear Programming in Operations Management • Formulation of Linear Programming Problems • Solution of Linear Programming Problems (Modified) • The Transportation Problem in Linear Programming

SOLUTION OF LINEAR PROGRAMMING PROBLEMS

Graphical Method

Added under ‘Graphical Method’ in page 45 of the first edition) Problem The plant manager of a plastic pipe manufacturer has the opportunity of using two different routings for a particular type of plastic pipe. Routing 1 uses Extruder A and routing 2 uses Extruder B. Both the routings require the same melting process. The following table shows the time requirements and capacities of these processes. Time Requirements (hr/100ft)

Process Routing 1 Routing 2 Capacity (hr) Melting 1 1 45

Extruder A 1 0 80

Extruder B 0 1 120

Each 100ft of pipe processed on routing 1 uses 3kg of raw material, whereas that on routing 2 uses 2kg of raw material. The difference results from differing scrap rates of the extruding machines. Consequently, the profit per 100ft of pipe processed on routing 1 is Rs.300 and on routing 2 is Rs.400. A total of 180 kg of raw material is available.

a. Create a set of linear equations to describe the objective function and the constraints.

b. Use graphic analysis to find the visual solution

c. What is the maximum profit?

Solution

a) As the objective is to maximize profit, the objective function for this problem would be

Max Z = 300x + 400y

where x and y are decision variables representing the length of pipe produced using routing 1 and routing 2 respectively.

In the melting process, a maximum of 45 hrs is available for both the routings. Hence,


18

x(1) + y(1) ≤ 45

x + y ≤ 45

In routing 1, Extruder A can use a maximum of 80hrs, which means

x(1) + y(0) ≤ 80

x ≤ 80

Similarly, in routing 2, Extruder B can use a maximum of 120 hrs leading to

x(0) + y(1) ≤ 120

y ≤ 120

The available raw material is 200kg and routing 1 consumes 3kg while routing 2 consumes 2 kg for every 100ft. Thus, 3x + 2y ≤ 180

Hence, the objective function is Max Z = 300x + 400y The constraints are x + y ≤ 45 x ≤ 80 y ≤ 120 3x + 2y ≤ 180 where, x ≥ 0 and y ≥ 0 b) To get the visual solution, the data has to be plotted on a graph. To plot the graph, the inequalities in the constraints have to be converted into equalities. Thus, x + y = 45 x = 80 y = 120 3x + 2y = 180 In x + y = 45, when x = 0, y = 45 and when y = 0, x = 45 In 3x + 2y = 180, when x = 0, y = 90, and when y = 0, x = 60 c) To find the maximum profit, the coordinates of the corner points of the feasible region have to be substituted in the objective function. Corner point A (0,100) 300 x 0 + 400 x100 = 40,000 Corner point B (60,0) 300 x 60 + 400 x 0 = 18,000 Corner point C (80,0) 300 x 80 + 400 x 0 = 24,000 Corner point S (80,120) 300 x 80 + 400 x 120 = 24,000 + 48,000 = 72,000 Corner point D (0, 100) 300 x 0 + 400 x 100 = 40,000 Hence, Corner point S gives the maximum profit of Rs.72,000.

Allocating Resources to Strategic Alternatives

19

ERRATA

Graphical Method

In Figure 4.1 in page 44 of the first edition, the optimum solution at the point B was not correctly represented. The graphical method explains the process of obtaining a solution of a linear programming problem in a simple way. The procedure can be explained in the following steps:

Step 1: Formulate the linear programming problem by identifying the decision variables, the objective function and the constraints.

Step 2: Convert the inequality constraints to their equalities and plot them on a graph (in linear form).

Step 3: Using the inequalities in each constraint, determine the feasible region.

Step 4: Write down the corner points of the solution area. Substitute the values in the objective function. The optimum solution is obtained at any of these points.

The feasible solution area for the problem formulated in the Table 4.3 is shown as OABC in the Figure 4.1. From the figure, we can understand that the optimum solution is at the point (2000/3, 5000/3).

C B

A

20

Y

O (0,0)

20

60

80

40 80 X

Figure 4.2: Graphical Method

60

x + y = 45 40

3x + 2y = 180

S 120

100 120

Feasible region

x = 80

y=120

100

D


20

Figure 4.1: Optimum Solution (Graphical Method)

Facility Location and Layout • Importance of Location • Factors Affecting the Location Decisions • General Steps in Location Selection and Location Decision Process • Location Evaluation Methods (Modified) • Facility Layout • Basic Layout Formats • Developing a Process layout • Developing a Product Layout (Modified) • Developing a Cellular Manufacturing Layout • Japanese Approaches and Trends in Manufacturing Layouts • Service Facility Layouts

LOCATION EVALUATION METHODS

Cost-Profit-Volume or Break-Even Analysis

(Added under ‘Cost-Profit-Volume or Break-Even Analysis’ in page 83 of the first edition)

Problem Atlas Ancillaries plans to set up a manufacturing unit for ‘100 HP’ motors. The management has identified Gurgaon and Delhi as the potential areas to set up the unit. The fixed costs per year and the variable costs per unit for Gurgaon are Rs.6,00,000 and Rs.750 respectively. The costs for Delhi are Rs.7,50,000 and Rs.650. Each motor is priced at Rs.1500 and the expected sales are 9000 units per year. Which of the two locations will be more profitable for Atlas? Solution Let us calculate the total costs (sum of the fixed and variable costs) at both the locations when 9000 units of goods are sold. Total cost at Gurgaon = Rs.6,00,000 + (750 × 9000) = Rs.127,50,000 Total cost at Delhi = Rs.7,50,000 + (650 × 9000) = Rs.133,50,000 Total sales revenues of the firm per year = 1500 × 9000 = Rs.135,00,000. Therefore, the profits of the company if they were set up in the given locations would be as follows:

Profit at Gurgaon = Rs.135,00,000 − Rs.127,50,000 = Rs.750,000

Profit at Delhi = Rs.135,00,000 − Rs.133,50,000 = Rs.150,000


22

From these calculations, it is clear that Gurgaon will the more profitable location to set up the new plant for producing 9000 motors per year.

DEVELOPING A PRODUCT LAYOUT

Line Balancing

Steps in assembly line balancing

Added under ‘Steps in assembly line balancing’ in page 90 of the first edition) The following steps are needed to balance an assembly line: i. The sequential relationship among different tasks is specified by using a precedence diagram. The cycle time is determined by using the following formula:

dayper output Required

dayper timeProduction timeCycle =

ii. The theoretical minimum number of workstations required to satisfy the cycle time is determined using the following formula:

CT

tN =

Where Nt = Theoretical number of workstations T = Sum of task times C = Cycle time iii. A set of rules is identified to shortlist and select the tasks to be assigned to workstations. A sample set of rules is given below. a) Identification of feasible (remaining) tasks for the same station: From the unassigned tasks, identify the task(s) which can be assigned next to the same station, subject to two constraints: • The precedence rules should not be violated. • The individual time required for each of these feasible (remaining) tasks should

be less than the unassigned time for the station, where Unassigned time for a station = Cycle time – (Sum of the time required for all previous tasks that have been assigned to the station) Note: • When there is no feasible (remaining) task for the same station, move on to the

next station. • When there is exactly one feasible remaining task for the same station, assign it

as the next task for the same station. • When there are multiple feasible remaining tasks for the same station, use the

following tiebreaker rules to shortlist/select the next task for the same station. b) Shortlist the tasks with most followers, among the feasible (remaining) tasks for the same station:

Facility Location and Layout

23

From the feasible (remaining) tasks for the same station, shortlist the task(s) which has (have) the most followers.

c) Select the task with the longest operation time:

From the shortlisted tasks with most followers, select the task which has the longest operation time, and assign it as the next task for the same station.

Sometimes, there may be multiple such tasks. In this case, one of these tasks with the longest operation time can be (arbitrarily) assigned as the next task for the same station.

iv. This set of rules is applied iteratively till all the tasks are assigned. At the end of this process, the actual number of work stations (Na) required may be greater than or equal to the theoretical number of work stations (Nt).

The efficiency of the balance is calculated by using the following formula.

CaNTEfficiency×

=

Where, T = Sum of task times

Na = Actual number of workstations C = Cycle time

v. The balance is accepted if the efficiency is satisfactory, otherwise balancing is done using a different decision rule.

ERRATA LOCATION EVALUATION METHODS

Cost-Profit-Volume or Break-Even Analysis

In the Figure 6.1 in page 81 of the first edition, TCA (total cost at location A) and TCB (total cost at location B) were incorrectly represented.

Figure 6.2: Higher Cost Locations Providing Higher Profits


24

DEVELOPING A PRODUCT LAYOUT

Line Balancing

Steps in assembly line balancing

(Problem 3 in page 91 of the first edition has been reworked as follows)

Problem The desired daily output for an assembly line is 500 units. The assembly line operates for a period of 420 minutes a day. The process involves the tasks A, B, C, D, E, F, G, H, I, J and K. Balance the assembly line and calculate the cycle time and efficiency of the assembly line.

Task Task Time (seconds) Tasks that must precede A 45 - B 11 A C 9 B D 50 - E 15 D F 12 C G 12 C H 12 E I 12 E J 8 F,G,H,I K 9 J 195

Solution


25

Following is the precedence diagram of all the tasks. The time required (in seconds) for completion of each of the tasks is given below.

Given that the operation time per day is 420 min, i.e. (420×60) sec. The sum of the task times of the process, T = (45 + 11 + 9 +50 + 15 + 12 + 12 +12 + 12 + 8 + 9) = 195 sec.

dayper output dayper timeOperation timeCycle =

500

60420×=

= 50.4 sec. Theoretical minimum number of workstations required,

50.4195

CTN t ==

= 3.87 Therefore, a minimum of 4 workstations are required to balance the assembly line. We arrange tasks in order of the largest number of following tasks.

A

B C F

G

J D E H

I

K

11

45 sec

50 sec

9 sec 12 sec

15 sec

12 sec

12 sec

12 sec

8 sec 9 sec


26

Now, we assign tasks as shown in the table below, following the rules given below:

A) Identification of feasible (remaining) tasks for the same station: From the unassigned tasks, identify the task(s) which can be assigned next to the same station, subject to two constraints:

a. The precedence rules should not be violated.

b. The individual time required for each of these feasible (remaining) tasks should be less than the unassigned time for the station, where

Unassigned time for a station = Cycle time – (Sum of the time required for all previous tasks that have been assigned to the station)

Note:

When there is no feasible (remaining) task for the same station, move on to the next station.

When there is exactly one feasible remaining task for the same station, assign it as the next task for the same station.

When there are multiple feasible remaining tasks for the same station, use the following tiebreaker rules to shortlist/select the next task for the same station.

B) Shortlist the tasks with most followers, among the feasible (remaining) tasks for the same station:

From the feasible (remaining) tasks for the same station, shortlist the task(s) which has (have) the most followers.

C) Select the task with the longest operation time: From the shortlisted tasks with most followers, select the task which has the longest operation time, and assign it as the next task for the same station. Sometimes, there may be multiple such tasks. In this case, one of these tasks with the longest operation time can be (arbitrarily) assigned as the next task for the same station.

Application of the above rules for balancing the assembly line given in the problem: To begin with, no tasks are assigned to any station. So, the unassigned time for Station 1 is 50.4 seconds. Subject to the precedence rules and the time constraints, A or D can be considered as the feasible remaining tasks. Since A has the most followers, we assign Task A to Station 1, as shown below.

Task Number of Following Tasks

A 6 B or D 5

C or E 4

F, G, H, I 2

J 1

K 0


27

Station Task Time Unassigned Time

Feasible remaining tasks (for the same station)

Task with most followers

Task with longest operation time

Station 1 - 0 50.4 A, D A A Station 1 A 45 5.4 None None None

Now, the unassigned time for Station 1 is only 5.4 seconds and neither of the potential next task (B or D) can be assigned next to this station, because of the time constraint. So we move on to Station 2. Following the same rules described above, the entire table can be filled as shown below.

Station Task Time Unassigned Time

Feasible remaining tasks (for the same station)

Task with most followers

Task with longest operation time

- 0 50.4 A, D A A Station 1 A 45 5.4 None None None

- 0 50.4 B, D B, D D Station 2 D 50 0.4 None None None

- 0 50.4 B, E B B B 11 39.4 C,E C,E E E 15 24.4 C,H,I C C C 9 15.4 F,G,H,I F,G,H,I F,G,H,I

Station 3

F 12 3.4 None None None - 0 50.4 G,H,I G,H,I G,H,I G 12 38.4 H,I,J H,I H,I H 12 26.4 I,J I I I 12 14.4 J J J

Station 4

J 8 6.4 None None None - 0 50.4 K K K Station 5 # K 9 41.4 All tasks have been assigned.

# The last task (Task K) could not be assigned to Station 4, since the unassigned time (6.4 seconds) was less than the time required for Task K. So, we need a fifth workstation to perform Task K. Therefore, actual number of workstations, Na = 5

Efficiency of the assembly line CN

T

a=

50.45195×

= = 0.77

Aggregate Planning and Capacity Planning

• Overview of Planning Activities • The Aggregate Planning Process • Strategies for Developing Aggregate Plans (Modified) • Aggregate Planning Techniques • Master Production Schedule (Modified) • Implementing Aggregate Plans and Master Schedules • Capacity Planning (Modified)

STRATEGIES FOR DEVELOPING AGGREGATE PLANS

Pure Planning Strategies

Varying the size of inventory: (Added after Problem 2 in page 104 of the first edition) Problem Aggregate demand for product X for the next four months is given in the following table:

Jun Jul Aug Sept Demand 5000 4600 5200 4800 Working Days 23 24 22 23

Given: Opening stock of inventory = 500 units Inventory holding cost = Rs. 40 Worker productivity = 20 units/day

Worker strength = 10

Shortage cost (due to lost sales) = Rs. 30/unit

Based on the above information, generate a production plan with varying inventory levels.

Solution The aggregate production plan with varying inventory is given in Table 9.2.

In aggregate planning with varying inventory level, the workforce is kept constant.

Actual production in a month is calculated as = (Number of working days) × (Number of workers) × (worker productivity in units/day)


29

Table: 9.2: Production Plan with Varying Inventory Level Jun July Aug Sep

Opening stock of inventory 500 100 300 0 Working days 23 24 22 23 Actual Production 4600 4800 4400 4600 Demand Forecast 5000 4600 5200 4800 Shortage in Supply (unmet demand)

0 0 500 200

Shortage Cost (due to lost sales)

0 0 15000 6000

Safety Stock 0 0 0 0 Closing Inventory 100 300 0 0 Inventory carrying costs 4000 12000 0 0

Case 1: If (Opening inventory + actual production) >= Demand forecast, then a. Closing inventory = Opening inventory + actual production – Demand

forecast

b. Shortage in supply = 0

Case 2: If (Opening inventory + actual production) < Demand forecast, then a. Shortage in supply = Demand forecast – (Opening inventory + actual

production)

b. Closing inventory = 0

Shortage cost (due to lost sales) = (Units short) × (Per unit shortage costs)

Inventory carrying costs = (Closing inventory) × (Per unit inventory holding costs)

Closing inventory in one month is taken as the opening inventory for the next month.

MASTER PRODUCTION SCHEDULE

Master Schedule Formation

Make-to-order items (Added after Problem 3 in page 112 of the first edition) Problem Develop a master production schedule for the following data:

Forecasted demand for the next 8 weeks: 24, 25, 23, 26, 25, 27, 26, 27

Orders booked: 28, 26, 22, 22, 23, 25, 29, 27

Inventory on hand = 30

Lead time = 1 week

Production lot size = 50 units

Solution First week: Requirement for the first week based on the existing order is 28 units (given). This requirement can be satisfied by using the on-hand inventory.


30

Projected inventory on hand at the end of first week is = On-hand inventory + MPS quantity - projected requirements for the week = 30 + 0 - 28 = 2 Second week: Forecast for second week is 25 units but the orders received are for 26 units. Inventory on hand at the end of first week is 2 units, which is not sufficient to satisfy the second week’s requirements. So to make up for the deficiency, organization schedules for MPS quantities. As the lead time is one week, the production should commence in the first week itself to satisfy the requirements of the second week. As can be seen from Table 9.4, MPS start quantity for first week is 50 units (production lot size). Now at the end of second week projected inventory on hand = On hand inventory + MPS quantity - Projected requirements for the week = 2 + 50 - 26 = 26 units Similarly, projected inventory at the end of each week and MPS quantities is calculated.

Table 9.4: MPS for 8 weeks

1 2 3 4 5 6 7 8 Forecast 24 25 23 26 25 27 26 27 Orders 28 26 22 22 23 25 29 27 MPS quantity 0 50 0 50 0 50 50 0 Projected on-hand inventory

2 26 3 27 2 25 46 19

MPS start 50 50 50 50 The company has to schedule production in the first, third, fifth, and sixth weeks.

CAPACITY PLANNING

(Added under ‘Capacity Planning’ in page 113 of the first edition)

Measuring Capacity Capacity can be measured in terms of the inputs or outputs of the conversion process. For a company with single product or similar products in its portfolio, capacity is given as the number of units per month. But for an organization with a multitude of products in its product portfolio, capacity is measured in a common unit such as sales in a monetary unit per week, month, or year. In service organizations measuring output is not possible, so the input rate capacity is used to measure the capacity. For example, a hospital measures capacity in terms of the number of beds available, airline use the number of seats available in a particular time period, etc. Measuring available capacity exactly is not always possible as capacity is dependent on many factors. For example, machine breakdown, employee absence, and other similar factors can lead to disruption in production, affecting the final capacity. So, decisions on capacity availability are taken after considering unavoidable interruption in productions.

To measure capacity, we use the following formula:

Capacity = Available time × Utilization × Efficiency


31

The capacity utilization rate measures the capacity level at which a production process is operating.

100avaialableCapacity

usedCapacity raten utilizatioCapacity ×=

Capacity available indicates the capacity level for which the process is designed (i.e. designed capacity). Capacity utilization rate is measured in percentage terms.

Determining Capacity Requirements Future capacity requirements need to be accurately determined to ensure good performance in the future. Required capacity is determined on the basis of the specifications of individual product lines, production capabilities in existing facilities, etc. The steps for determining capacity requirements are as follows: 1. Employ forecasting methods and find the individual product demand within the

planning horizon. 2. Identify the equipment and human resource requirements to meet the forecasted

demand. In case of multiple products or services, identify the time required for switching from one product or service to another.

3. Compare capacity required with available capacity for the required time period and identify the gaps.

4. As determining exact capacity requirements is difficult, organizations in general allocate extra capacity to meet any contingencies in the future. The extra capacity allocated for uncertain future requirements is known as the capacity cushion.

Economies of Scale The basic theory behind the economies of scale is that as the size of an operation increases, per unit cost of production decreases. The reasons include decrease in the fixed costs of production per unit of output, and adoption of efficient process and technologies like automation, which are not economically feasible for small-scale operations. But when the firm expands beyond a point, diseconomies of scale become apparent, because the expenditure incurred for maintaining large operations becomes uneconomical.

Diseconomies of scale

Economies of scale

O

Production volume (units)

Average unit cost of

output (Rs.)


32

The reasons for diseconomies include increased distribution and storage costs, complexities in operations, and high costs of modification and replacement of existing facilities. So, an organization has to determine capacity required to achieve economies of scale. In Figure 9.6, the relationship between the production level and the average unit cost is shown. The resulting economies/diseconomies of scale are apparent. Point O shows the optimum level of output where an organization can take full advantage of economies of scales without facing diseconomies of scale.

Planning Service Capacity

In a service organization, planning for capacity should take into consideration when and where the capacity is needed. Services can neither be produced in anticipation of demand nor stored as inventories. The service capacity should be located near the customers because most service delivery process involves customers. Unlike manufacturing organizations, a service organization based in one geographical location usually cannot efficiently serve customers in other geographical area. As a result, service organizations generally operate through branches or franchises because of the nature of the services, which involves customer participation in the service delivery process. As customers cannot come to single location to avail services, services organizations tend to increase capacity by establishing new branches and facilities. There are few exceptions to these rules; many educational institutions, hospitals, holiday resorts, etc. tend to concentrate on increasing capacities in one location itself. Further with the expansion of the Internet, many retailers are taking up the option of providing service online from a single location rather than establishing brick and mortar outlets.

Fundamentals of Inventory Control

• Purpose of Inventories • Inventory Costs • Inventory Systems

• Economic Order Quantity Model (Modified) • Inventory Classification Models (New)

ECONOMIC ORDER QUANTITY MODEL

Optimal Order Quantity (Added after Problem 1 in page 126 of the first edition) Problem

A telecom distributor stocks a certain kind of electrical switch relay system that costs Rs.50 per unit and has an annual demand of 10,000 units. The ordering costs are estimated at Rs.80 per order and carrying costs are estimated at 20% of the cost of the item. Calculate the EOQ.

Solution:

Given :

Cost of the switch = Rs.50

Demand = 10,000 units

Ordering cost = Rs.80 per order

Carrying cost = (20/100)50 = 0.2 x 50 = Rs.10

EOQ=hC

Do2C=

1000010802 ×× = 400 units

So the optimal order quantity for the firm is 400 units. The firm would be able to minimize total cost i.e. carrying and ordering cost if it orders in lots of 400 units.

INVENTORY CLASSIFICATIONS MODELS

(New Heading added after ‘Economic Quantity Order Model’ in page 126 of the first edition) To exercise proper control over the inventory items, organizations classify and categorize these items. This classification allows operations managers in devising control plans relevant and effective for items with similar attributes. These methods are sometimes called selective control methods.


34

ABC Classification ABC is one of the most widely used inventory classification models otherwise known as “Always Better Control” model. As per ABC classification, items are classified on the basis of their annual consumption value. In an organization, few items have maximum turnover in terms of annual consumption or usage value. Annual usage value of an item is given by the relationship

Annual usage value = annual requirement × per unit cost

Category ‘A’ includes items that account for approximately 10% of the total inventory items and 75% of the inventory investment. These items need high level of control.

Category ‘B’ includes items that account for approximately 15% of total inventory items and inventory investments.

Category ‘C’ includes items that account for approximately 75% of the total inventory items with approximately 10% of the total inventory investment. The ABC classification method is described in detail in Chapter 12.

VED Classification

VED classification is based on the importance of a particular item in the production process. Under VED, the items, which are critical for production are classified as V (vital), the items, which are important in the production process but not critical are classified as E (essential) and the items, which consists of non essential and do not influence the production process are classified as D (desirable). Items categorized as V need maximum control and investment. V items may not be the most expensive but are critical to the production process. ABC classification of items is based on the annual usage value whereas VED method is based on the criticality of the item in the production process.

FSND Classification

Under FSND classification, goods are classified on the basis of their turnover. F items are fast moving, S items are slow moving and N items are non-moving items and D items are dead or the items which have not been issued for few years. In most cases, the items identified as D are normally sent for disposal. The classifications of items as FSND are based on the number of issues or withdrawal of items from stores over a period of time. This period is decided based on the kind of inventory a firm is holding. Time period for the movement of inventory in a steel factory may be different from that of an electronics firm.

Purchase Management • Importance of Purchasing • Organizing Purchasing • Responsibilities of a Purchase Manager • Purchasing Process • Duties of Buyers • Make-or-Buy Decisions (Modified) • Ethics in Buying

MAKE-OR-BUY DECISIONS

(Added under ‘Make or buy Decisions’ in page 136 of the first edition)

Problem 1

Raj Metals can purchase brass plates for Rs.30 per unit or produce it in-house where the associated fixed and variable costs are Rs.50,000 and Rs.20 respectively. If the demand for this component is 5000 units per annum, ascertain whether it is economical to make the product in-house rather than sourcing it.

Solution

Given:

Demand = 5000 units

F = Rs. 50,000

V = Rs. 20

P = Rs.30

Q = 5000

If the item is made in-house, the organization incurs a fixed cost, say F, on installing the necessary equipment and facilities. It also incurs a variable production cost, which is equal to the unit variable cost (V) times the number of units demanded (Q).

Total Cost to manufacture = (VQ) + F

= (23 x 5000) + 50000

= 115,000 + 50000 = 165,000

The total cost of buying is the product of price per unit (P) and the number of units procured (Q), i.e. Total Cost if purchasing = P × Q

= 5000 x 30 = 150,000

As the cost to produce the item is more than the cost of purchasing it, it is not economical for the company to produce the component in-house.

Problem 2

An automobile company plans to utilize its existing manufacturing facility to produce gears that the company is now buying for Rs.250 per unit. If the company produces the gears, it will incur materials costs of Rs.75, labor costs of Rs.100, and overhead


36

costs of Rs.25 per unit. The annual fixed cost associated with the unused capacity is Rs.200,000. Demand over the next year is estimated at 25,000 units. Would it be profitable for the company to make the gears?

Solution Demand (Q) = 25,000 units Variable cost (V) = material costs + labor costs + overhead costs = 75 + 100 + 25 = 200 Fixed cost (F) = Rs. 200,000 Price per unit (P) = Rs. 250 Total cost to purchase = P × Q = Rs.250 x 25000 = 6,250,000 Total cost to produce = Fixed cost + (Variable cost x demand) = 200,000 + (200 x 25000) = 200,000 + 5,000,000 = 5,200,000 As the total cost of producing the gears is less than the total cost of purchasing them by Rs.10,50,000, the firm can produce the gears in-house.

Materials Management • Necessity of Materials Management • Functions of Materials Management • Materials Management Technology • Materials Management Techniques (Modified)

ERRATA

MATERIALS MANAGEMENT TECHNIQUES

Kanban Systems Figure 13.3 in page 183 of the first edition was incorrectly represented

A dual-card Kanban system

The dual-card Kanban system makes use of two Kanban cards. The two Kanbans are a conveyance card and a vendor card. In a single-card Kanban system, we assume that the required quantity of inventory is available in the inventory department. In the dual card system, the required inventory is obtained from a vendor and a vendor authorization Kanban is used in the transactions.

The functioning of a dual card system is shown in Figure 12.3 and can be described in the following steps:

• The process of working of a conveyance card in this system is similar to that in the single-card system. A conveyance Kanban is put in an empty tray at point A and a material handling agent moves it to point B in the inventory department. The tray is collected at point C and sent to point D in the assembling area similar to the single card system.

• In this system, the vendor Kanban is introduced at point X, authorizing a vendor (at point Y) to deliver the materials that are specified in the card.

• After receiving the card, the vendor delivers the materials into the empty trays that are available at point Z.

• The filled container is placed in the bins at the position X. It remains at this position till a conveyance card arrives from Point B to Point C, authorizing the movement of material from Point C to Point D.

• Once the container is authorized to move, the vendor Kanban is removed and sent to the vendor and cycle repeats itself


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Figure 12.3: Dual-Card Kanban System

Source: Sang M Lee and Marc J Schniederjans, Operations Management (Chennai: All India Publishers and Distributors, 1997) 258. .

Materials Requirement Planning • Fundamentals of Materials Requirement Planning • Components of an MRP System (Modified) • Advantages and Disadvantages of an MRP System • Problems in Implementing MRP Systems • Manufacturing Resource Planning

COMPONENTS OF AN MRP SYSTEM

MRP System Inputs

Bills of material (Added after Problem 1 under ‘Bill of Material’ in page 194 of the first edition) Problem Given the product structure tree shown in Figure 13.4, how many units of F overall are required to manufacture 200 units of A? The on-hand inventory available is 100 units of subassembly D and 100 units of subassembly B.

Solution Quantity of A to be produced = 200 units Quantity of D required to produce 200 units of A = 400 units Available inventory of D = 100 units. Quantity of D to be produced = 300 units

A

B (3) C (1) D (2)

E (1) F (3) H (4) I (2) J (1) F (4)

G (2) F (1) G (3)

Figure 13.4: Bill of Material for Product A


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4 units of F are required to produce one unit of D. Hence, 1200 units are required to produce 300 units of D. Quantity of B required to produce 200 units of A = 600 units Available inventory for B = 100 units Quantity of B to be produced = 500 units 3 units of F are required to produce one unit of B. Hence, 1500 units are required to produce 500 units of B. For every unit of A, one unit of C is required and for every unit of C, two units of F are required. Hence, for 200 units of C, 400 units of F are required. The total quantity of F required to produce 200 units of A = 1200 + 1500 + 400 = 3100 units.

MRP System Information Processing

Offsetting (Added after Problem 2 under ‘Offsetting’ in page 196 of the first edition) Problem The BOM for product X is given in Figure 13.6. Develop an MRP schedule for items P, Q, R, and S so that 300 units of X can be produced in the 6th week. On-hand inventory of component Q is 100 units.

Solution From the given data, the MRP schedule for items P, Q, R, and S are as shown in the table 13.3. Here, the lead times for X = 2 weeks, P = 2 weeks, Q = 1 week, R = 1 week, and S = 2 weeks

Table 13.3: MRP Schedule for Item P, Q, R, and S Week Number 1 2 3 4 5 6

Item: P Gross requirements 600

Quantity =2 Quantity on Hand 0

LT =2 Net requirements 600

P (2) Q (2)

X

R (2) LT=1

LT=2 LT=1

Figure 13.6: BOM for Product X

S (3) LT=2

LT=2

Materials Requirement Planning

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Week Number 1 2 3 4 5 6

Lot Size =1 Planned-order receipts 600

Planned-order releases 600

Item: R Gross requirements 1200





Item: Q Gross requirements 600





Item: S Gross requirements 1500





From the BOM file, we can see that 2 units of P are required for producing 1 unit of X. Thus 300 units of X would require 600 units of P in the sixth week. Similarly, 2 units of Q are required to produce 1 unit of X. Hence, 300 units of X would require 600 units of Q in the sixth week. As the on-hand inventory of component Q is 100 units, the net requirement for Q is 500. The lead-time to acquire Q is 1 week, so 500 units planned-order release are placed in week 3. Similarly we can deduce the schedule for R and S as shown in Table 13.3.

Operations Scheduling • Purpose of Scheduling • Scheduling Methods (Modified) • Scheduling Activities • Scheduling by Type of Operations • Scheduling Personnel in Service Operations (Modified) • Scheduling Techniques (Modified)

SCHEDULING METHODS

Backward Scheduling

(Added after Problem 1 under ‘Backward Scheduling’ in page 35 of the first edition)

Problem

National Machines Limited has two job orders, X to produce 1000 nuts and Y to produce 500 screws. Both orders have to be processed on two machines 1 and 2. The route sheets for the jobs are given in the following table. Both the jobs should be ready in the next nine hours and both the machines are to start processing from now onward. Develop schedules for both the jobs using forward and backward scheduling.

Job X Route Sheet Job Y Route Sheet

Routing Sequence

Machine Processing Time

(Hours)

Routing Sequence

Machine Processing Time

(Hours)

1 1 1 1 2 2

2 2 2 2 1 2

3 1 2 3 2 2

Total 5 Total 6

Solution

Forward Scheduling:

Time 1 2 3 4 5 6 7 8 9

Machine 1 X1 Y2 Y2 X3 X3

Machine 2 Y1 Y1 X2 X2 Y3 Y3

Here, 1000 nuts along with 500 screws can be produced using the given sequence and machines earliest by the end of 6th hour.

Operations Scheduling

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Backward Scheduling:

Time 1 2 3 4 5 6 7 8 9

Machine 1 X1 Y2 Y2 X3 X3

Machine 2 Y1 Y1 X2 X2 Y3 Y3

Here, the job of producing 1000 nuts can start latest in the 5th hour, and for 500 screws using the given sequence and machines by the 4th hour so as to finish by the end of 9 hours.

SCHEDULING PERSONNEL IN SERVICE OPERATIONS

(Added after ‘Scheduling by Type of Operations’ in page 242 of the first edition) Service firms have to develop weekly, daily and hourly personnel schedules. Some of the approaches that firms can follow in this regard are discussed below.

Scheduling Consecutive Days Off A weekly schedule for each employee two consecutive days off in a week can be drawn up as shown in the following example.

Suppose an educational institute wants to develop a schedule for its teaching faculty. The table gives the number of faculty required on each day of the week.

Monday 3

Tuesday 4

Wednesday 4

Thursday 2

Friday 2

Saturday 2

Sunday 1

Now we assign each faculty consecutive days off using the method given below.

1. Assign Faculty A to all the days that require staffing. 2. Then identify the two consecutive days with the lowest numbers (lowest pair) and

these days become off for Faculty A. The lowest pair is the pair in which the highest number of the pair is less than or equal to the highest number of any other pair. The numbers against Sunday and Monday are also considered as a pair.

3. If a tie occurs, choose the pair with the lowest number (or requirement) on an adjacent day (before or after the pair).

4. If a tie occurs again, select the first of the tied pairs. Faculty A:

Mon Tue Wed Thu Fri Sat Sun

3 4 4 2 2 2 1

So Faculty A is assigned offs on Saturday and Sunday, as the lowest pair is (2, 1).

Now we assign the days off to the remaining faculty. The row for Faculty B is developed by subtracting 1 from all the days that Faculty A works.


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Faculty B:


2 3 3 1 1 2 1

Here, the lowest pair is (1, 1) and it exists on Thursday and Friday. So Faculty B gets offs on Thursday and Friday.

Now the row for Faculty C is,


1 2 2 1 1 1 0

Now there is a tie for the least pair. The pair is (1, 1) and it occurs on Thursday and Friday; as well as Friday and Saturday. We select Friday and Saturday as the lowest pair, as the adjacent value to this pair is the lowest.

So Faculty C is off on Friday and Saturday.

Since there is no requirement of faculty, Faculty C will also be off on Sunday.

Now the row for Faculty D is,


0 1 1 0 1 1 0

So, Faculty D is off on Sunday and Monday. He also gets off on Thursday as there is no requirement of faculty on that day.

So the Institute can take two employees, Faculty A and Faculty B on a permanent basis and take Faculty C and Faculty D on a part-time basis. Faculty A and Faculty B handle one session each and they can be engaged in the other activities of the institute like content development, research and placement activities.

Faculty C and Faculty D take 4 sessions every week on a part-time basis.

Scheduling Daily Work Times

One can also determine the least number of workers required to accomplish the daily work load.

This is explained in the following example. A firm is engaged in manufacturing four types of toys A, B, C and D. Each toy is manufactured through three functions: set up work, fabrication, and decorating work.

The following table provides the production rate (number of units per hour) of each function for each type of product.

Production Rate (units per hour) Product Vol. Setup work Fabrication Decorating work

A 100 3.3 2.5 5

B 75 2.5 1.5 2.5

C 150 2.5 1.5 3.75

D 200 5 2.5 5


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Based on the above information, we calculate the time required for completing each function of all types of products.

Processing times Product

Setup work Fabrication Decorating work

Total time

A 30 40 20 90

B 30 50 30 110

C 60 100 40 200

D 40 80 40 160

Total 160 270 130 560

By assuming each employee works for eight hours a day, we obtain the number of employees required by dividing the total number of hours with 8.

Therefore, the number of employees required for setup work is 160/8 = 20, fabrication work is 270/8 = 33.75, and decorating work is 130/8 = 16.25.

In total, 70 employees are required to complete all the work.

So, the firm should engage 20 employees for setup work, 33 employees for fabrication and 16 employees for decorating work; and one employee can be engaged in fabrication work for 75 percent of time (6 hours a day) and decorating work for 25 percent of time (2 hours a day).

Scheduling Hourly Work Times

Service requirements may vary from hour to hour in service firms such as hotels and restaurants. So these firms engage more workers when the demand is high. This type of personnel scheduling uses a rule called the ‘first hour principle.’ According to this principle, in the first period, the number of workers required in that period is assigned. Subsequently, additional workers are assigned as and when required. This approach is explained in the following example.

A hotel works from 5 a.m. to 11 p.m. everyday and the requirements of the workers for each 2- hour period is given below.

5 a.m. – 7 a.m. 8

7 a.m. – 9 a.m. 14

9 a.m. – 11 a.m. 12

11 a.m. – 1 p.m. 18

1 p.m. – 3 p.m. 16

3 p.m. – 5 p.m. 12

5 p.m. – 7 p.m. 12

7 p.m. – 9 p.m. 16

9 p.m. – 11 p.m. 14


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At 5 a.m., we assign 8 workers to work from 5 a.m. to 1 p.m.

At 7 a.m., we assign 6 new workers to work from 7 a.m. to 3 p.m.

Since the requirement of workers came down, we do not assign any new workers at 9 a.m. Therefore, 2 workers will remain idle for the period 9 a.m. – 11 a.m.

Again at 11 a.m., we assign 4 new workers to work from 11 a.m. to 7 p.m.

At 1 p.m., we assign 6 new workers, as 8 workers who stated working from 5 a.m. will leave by 1 p.m. These people will work from 1 p.m. to 9 p.m.

At 3 p.m., we assign 2 new workers to meet the requirement of 12 workers, as 6 workers among the present 16 workers will leave by 3 p.m.

Since there is no further increase in the requirement of workers, we do not assign any new workers at 5 p.m.



Therefore, the management of the hotel can recruit (8+6+0+4+6+2+0) = 26 workers on a permanent basis and 12 workers on a part-time basis. Of the 12 part-time workers, 8 workers work from 7 p.m. to 11 p.m. and 4 workers work from 9 p.m. to 11 p.m. The above description is summarized in Table below.

Time period Workers required

Newly assigned

On duty

Workers who leave at

the end of the period

Workers left

5 a.m. – 7 a.m. 8 8 8 - 8

7 a.m. – 9 a.m. 14 6 14 - 14

9 a.m. – 11 a.m. 12 0 14 - 14

11 a.m. – 1 p.m. 18 4 18 8 10

1 p.m. – 3 p.m. 16 6 16 6 10

3 p.m. – 5 p.m. 12 2 12 - 12

5 p.m. – 7 p.m. 12 0 12 4 8

7 p.m. – 9 p.m. 16 8 16 6 10

9 p.m. – 11 p.m. 14 4 14 2 12*

* Part time workers.

The management of the hotel can also opt for splitting of shifts. In this example, two workers remain idle from 9 a.m. to 11 a.m. Therefore, the management of the hotel can assign two workers to work from 5 a.m. to 9 a.m. and again from 7 p.m. – 11 p.m. If the management opts for splitting of shifts, it can reduce the number of part time workers to 10. The following table shows scheduling of the personnel if the management opts for splitting of shifts.


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Time period Workers required

Newly assigned

On duty Workers who leave at the end

of the period

Workers left

5 a.m. – 7 a.m. 8 8 8 - 8

7 a.m. – 9 a.m. 14 6 14 2+ 12

9 a.m. – 11 a.m. 12 0 12 - 12

11 a.m. – 1 p.m. 18 6 18 6 12

1 p.m. – 3 p.m. 16 4 16 6 10

3 p.m. – 5 p.m. 12 2 12 - 12

5 p.m. – 7 p.m. 12 0 12 6 6

7 p.m. – 9 p.m. 16 8+2 16 4 12

9 p.m. – 11 p.m. 14 2 14 4 10*

+ These workers will again join the work at 7 p.m. * Part time workers

SCHEDULING TECHNIQUES

Johnson’s Job Sequencing Rules

Johnson’s rule for n-jobs and m-machines (Added after Problem 5 in page 250 of the first edition)

Problem

Determine the optimum sequence of processing for the given sequencing problem of four jobs: A, B, C, and D and five machines: P, Q, R, S, and T. Also find the total elapsed time and idle time on each machine.

MachinesJob P Q R S T

A 9 3 3 5 10

B 12 6 6 7 7

C 7 4 5 2 8

D 13 7 3 4 9 Solution

Here, minimum time on P = 7 hrs,

Maximum time on Q, R, and S are 7, 6, and 7 hrs respectively.

Since the minimum on machine P is equal to and less than the maximum time on machines Q, R, and S respectively, the problem can move further.


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Now we take into account two fictitious machines, G and H, whose processing times are as follows

Job G = P + Q + R + S H = Q + R + S + T

A 20 21

B 31 26

C 18 19

D 27 23 Now we consider this as an n-job, 2-machine problem.

The optimum sequence is obtained using the steps given by Johnson and Bellman (mentioned earlier) as:

Job G = P + Q + R + S H = Q + R + S + T A 20 21 B 31 26 C 18 19 D 27 23

The calculation of total elapsed time is shown below.

Machine P Q R S T

Job C 0-7 7-11 11-16 16-18 18-26

Job D 7-20 20-27 27-30 30-34 34-43

Job B 20-32 32-38 38-44 44-51 51-58

Job A 32-41 41-44 44-47 51-56 58-68

The total elapsed time is 68 hrs.

The idle time of Machine P is, (68-41) = 27 hrs.

The idle time of Machine Q is, 7 + (20-11) + (32-27) + (41-38) + (68-44) = 48 hrs

The idle time of Machine R is, 11 + (27-16) + (38-30) + (68-47) = 51 hrs

The idle time of Machine S is, 16 + (30−18) + (44-34) + (68-56) = 50 hrs

The idle time of Machine T is, 18 + (34-26) + (51-43) = 34 hrs


Critical Ratio Method

(Added after Problem7 in page 251 of the first edition)

Problem

Dang Electronics has six different jobs in process with delivery requirements as shown in the table. Today is Day 60, and the company uses the critical-ratio scheduling technique. Rank the jobs according to priority. Which jobs have the highest and lowest priority?


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A B C D E Planned days 75 65 73 74 68 Work remaining in days 14 5 18 17 10

Solution

The critical ratios and the priority order are as shown in the following table.

Job Critical Ratio Priority Order

A (75-60)/14 =1.07 5

B (65-60)/5 = 1 4

C (73-60)/18 =0.72 1

D (74-60)/17= 0.82 3

E (68-60)/10 =0.80 2

From the table it can be observed that job C has the lowest critical ratio and hence the highest priority while job A has the highest critical ratio (ahead of schedule) and thus the lowest priority.

ERRATA


There has been a mistake in the table 17.2 in the first edition. The time out for Row 2 of Machine C is 61. As a result the Time in Row 3 of Machine C will also be 61. The Time out in Row 3 would be 75.

Johnson’s rule for three-stage production

Problem A firm is involved in five types of jobs, each of which must be processed on three machines, A, B and C in the order ABC. The processing time of each job (in hours) on the three machines is given below:

Job Processing Times

Ai Bi Ci

1 18 10 8

2 19 12 18

3 12 5 16

4 16 6 14

5 21 9 10

Determine the sequence for the five jobs that minimizes the total operation time. Also find the idle time of each machine; A, B and C.


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Solution First we check whether the problem satisfies the conditions discussed above. The smallest processing time on machine A is 12 and it is greater than or equal to the largest processing time on machine B, i.e. 12. Therefore, we can use Johnson's n-jobs, 3 - machine procedure to solve the problem.

Job Processing Times

Gi = Ai + Bi Hi = Bi + Ci

1 28 18

2 31 30

3 17 21

4 22 20

5 30 19

The optimum sequence is, 3, 2, 4, 5 and 1.

Now, we calculate the total operation time with the optimum sequence obtained.

Thus, the total operation time is 104 hours.

Idle time on Machine A = 104 – 86 = 18 hours.

Idle time on Machine B = 12 + [(31 – 17) + (47 –43) + (68 – 53) + (86 – 77)] + (104 – 96)

= 12 + 14 + 4 +15 + 9 + 8 = 62 hours.

Idle time on Machine C = 17 + (43 – 33) + (61 – 61) + (77 – 75) + (96 – 87)

= 17 + 10 + 0 + 2 + 9 = 38 hours.

MACHINE A MACHINE B MACHINE C Job Sequence

Time in

Processing Time

Time out

Time in

Processing Time

Time out

Time in

Processing Time

Time out

3 0 12 12 12 5 17 17 16 33

2 12 19 31 31 12 43 43 18 61

4 31 16 47 47 6 53 61 14 75

5 47 21 68 68 9 77 77 10 87

1 68 18 86 86 10 96 96 8 104


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Critical Ratio Method

Problem

Zenith Manufacturing Ltd. has started three jobs that require 28, 31 and 21 days to complete, respectively. The managing director of the company has engaged three teams to execute each of these jobs.

After the lapse of 18 days, the operations managers of the three jobs states that the times required to complete the jobs are 11, 14 and 8 days respectively.

Calculate the critical ratio of each job and find out which job is to be given priority.

Solution

The critical ratios and priorities are as follows:

Job Critical Ratio Priority Order

1 (28 − 18)/11 = 0.91 2

2 (31 − 18)/14 = 0.93 3

3 (21 − 18)/8 = 0.375 1

The lower the critical ratio, the higher the priority need in sequencing the job in the next day’s production activities. Here job 3 has the highest priority.

Productivity and Quality Management

• Productivity (New) • The Strategic Role of Quality • Role of Inspection in Quality Control • The Cost of Quality • Statistical Concepts in Quality Control (Modified) • Computers in Quality Control • Concept of TQM (Modified)

PRODUCTIVITY

(New heading added)

Productivity is a measure of the efficiency of an organization in terms of the ratio of the outputs to inputs. The higher the numerical value of this ratio, the greater the efficiency. Productivity measures the efficiency of the employees (teams or departments) in using the organization’s scarce resources to produce goods and services. It is an important tool for managers because it helps them track progress in terms of the efficient use of resources in producing goods and services. It helps them identify inefficient activities (if any) in the production process in terms of man-hours spent, material consumed, etc. Thus, productivity can be used as a controlling tool to ensure that all the resources are utilized judiciously and efficiently.

This section focuses on defining productivity, measuring productivity, and describing the factors that affect productivity in an organization.

Productivity Defined

The term ‘productivity’ was first used by French mathematician Quesnay in 1776. He defined it as the relation of output to input. In 1883, another Frenchman Littre defined productivity as the “faculty to produce”. In 1950, the Organization of European Economic Cooperation formally defined productivity as “the quotient obtained by dividing output by one of the factors of production. In this way, it is possible to speak of productivity of capital, investment, or raw materials, according to whether output is being considered in relation to capital, investment, or raw materials respectively.”

In general, productivity can be defined as the relationship between output from the system and inputs used to produce the output (products and services). In mathematical terms, it is the ratio of output to input.

InputOutputty Productivi =

Inputs

Inputs are all those factors of production like capital, men, material, machinery, etc. Inputs can be quantified into different categories like number of man-hours worked, quantity of material used, tons of raw material consumed, and power consumed in kilowatts (kW) or megawatts (MW). They can also be viewed in terms of costs associated like material costs, transportation costs, direct costs, overheads, etc. Input


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can also be categorized into tangible and intangible assets. Tangible assets include raw material consumed, components used, and machinery used, while intangible assets include leadership skills, knowledge of the workers, and training provided.

Outputs

Output may be in terms of the number of customers served in a restaurant, the number or volume of products produced in a factory, or the number of customer requests processed in a bank. It can also be tangible or intangible. The number of goods produced, tons of cement produced, etc., are tangible outputs while customer satisfaction is intangible. Operations managers need to develop means to quantify the intangible output to determine productivity accurately.

Components of Productivity Productivity can be improved in many ways such as by procuring quality raw material at the lowest possible cost, adopting an optimal mix of production factors, and training the workers. These different factors/components independently influence the productivity in different ways. The components can be categorized to include price efficiency, allocative efficiency, technical efficiency, and scale efficiency.

Price efficiency

Organizations always try to achieve a trade-off between the quality of the material purchased and the associated costs. Procuring good quality material at the lowest possible costs would reduce the overall cost of the product. Price efficiency comes in when such cost reductions are possible without compromising on the quality of the material purchased.

Allocative efficiency

Operations managers do consider an optimal mix of factors of production. For instance, a trade-off between capital and labor is necessary to maximize productivity due to constraints arising in utilizing either of them in a given situation. Hence, the adoption of an optimal mix of factors of production leads to allocative efficiency.

Technical efficiency

Productivity can be increased either by increasing the output with the same number of inputs or by producing the same output using fewer inputs. For instance, if 100 units are produced using 80 units of raw material, then productivity can be increased either by producing more than 100 units (say 110 units) by using the same number of inputs (80 units), or by producing 100 units by using less than 80 units of input (say 75 units of raw material). Either way, the productivity increases and this increase is termed as technical efficiency. Technical efficiency can be gained by modifying the existing production processes, introducing automation in production processes, etc.

Scale productivity

This component is concerned with the activity volume. The volume or size of operations has an impact on the productivity of the organization. For example, let us assume that the ordering cost for a raw material is fixed irrespective of the quantity ordered. In this situation, a small organization whose volume of output is low may find this ordering cost (input) more than a large organization, which has higher output volumes.


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Factors affecting Productivity

Productivity is one of the major concerns of managers as it helps organizations survive in a competitive environment. As productivity helps measure the efficiency and competitiveness of employees, departments, or organizations, it is considered an essential element in the control process. Although the need to improve productivity is felt in organizations, there is little consensus about the fundamental causes of low productivity and the ways in which they can be dealt with. Productivity in organizations is believed to be affected by several factors, some of which are mentioned here:

• Bottlenecks in the production process can hamper productivity. Poor layout design can be one of the reasons for bottlenecks. Further, productivity depends on the capacity of the slowest processing equipment. If steps are taken to improve the capacity of such equipment, it leads to a corresponding increase in the productivity.

• Training programs affect productivity. Rapid technological changes are taking place in organizations across the world. Therefore, employers have to train their employees to constantly upgrade their skills. However, the time (unproductive) spent by the employees in learning new technology reduces their productivity as the time spent on training eats into the total effective work time. Exhibit 18.1 details the National Productivity Council of India, a non-profit organization, which offers training and consulting to the firms on ways to improve productivity.

• The total production cycle time has an impact on total productivity. The production cycle time is the sum of the actual production time and idle time i.e. time spent by the material between processing. Hence, operations managers need to adopt proper scheduling in the plant to increase productivity by reducing idle time.

• The productivity of an organization may also suffer if social and legal obligations (for example, government policies) make it necessary for it to employ people without adequate skills.

• Workers often believe that their employers are exploiting them by increasing their workload from time to time on the pretext of improving productivity. Therefore, they resort to strikes, thereby leading to loss of productivity.

Exhibit 18.1

NPC and Productivity Improvement The National Productivity Council of India (NPC) was established by the Government of India in 1958 as a registered society. It is an autonomous and non-profit organization where equal interests are held by the government, the employers, and the workers’ organizations. NPC provides training and consultation to organizations in different industries on ways to improve productivity, apart from undertaking research in the area of productivity. In addition, it also actively implements the productivity promotion plans and programs of the Tokyo-based Asian Productivity Organization (APO).

NPC provides productivity training in various disciplines like Human Resource Management, Information Technology, Process Management, Technology Management, and Disaster Management. These training modules aim at improving the existing production methods and implementing new methods. For example,


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under the discipline of technology management, NPC provides training on spare parts management, maintenance management, etc. Team building, stress & time management, Six sigma work culture, etc. are the focus areas under Human Resource Management.

NPC offers consultation services to the industry, agriculture, and services sectors in a wide spectrum of domains. They include System/Industrial engineering, Agricultural productivity, Plant engineering, Energy management, Environment management, and Information technology. It also provides training and consultation services to firms under the Small Scale and Informal Sector (SIS) that includes associated government and non-governmental organizations.

Adapted from http://www.npcindia.org.

Measuring Productivity

Productivity can be measured in relation to a single factor (single factor productivity), a combination of factors (multifactor productivity), or all the factors taken together (total productivity). Single factor productivity and multifactor productivity can also be termed as partial productivity as all the factors of production are not considered while calculating productivity. Kendrick and Creamer defined partial productivity (single factor productivity) in 1965 as the ratio of gross or net output to one class of input. An example of single factor measurement of productivity is labor productivity, which typically measures output per unit of labor. Multifactor productivity takes into consideration more than one factor of production such as labor and materials. Total factor productivity includes all the factors of production (labor, materials, process, energy, and other inputs). According to Kendrick and Creamer, total factor productivity is the ratio of the real products originating in the economy, industry, or firm to the sum of the associated labor and capital (factor) inputs. In this section, we will focus on productivity measurement for tangibles (production) and intangibles (knowledge workers, services organizations).

Traditional approach to measurement of productivity Productivity is measured in terms of partial productivity and total productivity. As labor is one of the major sources of production costs for organizations, most productivity ratios are calculated by considering labor as the specific input. This partial productivity ratio is referred to as the labor productivity index or output per work-hour ratio.

(input)spenthourshours/manLabor(output)producedservicesand/orGoodstyproductiviLabor

−=

Material costs also affect productivity as they add up to 30% to 40% of the overall costs. Hence, operations managers also measure productivity in terms of material used.

(output)usedmaterialofQuantity(output)producedservicesand/orGoodstyproductiviMaterial =

In addition, managers often develop specific ratios that gauge productivity for particular outputs and inputs, such as sales per square foot of floor space, return on investment, amount of scrap for certain units of output, etc.

(input)usedcomponentsandmaterialrawofQuantity(output)producedservicesand/orGoodstyproductivirMultifacto =


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Many organizations measure productivity in terms of partial productivity (single factor or multi-factor). This is because it is difficult to measure total productivity due to the difficulty in identifying/understanding the particular input variable(s) (among many variables) that has led to lower productivity. Another problem with total productivity is that all the variables (inputs and outputs) must be expressed in the same units. But it is difficult to add the number of labor hours to the number of units of energy or any other units of an input.

[ ] ( )inputsmaterialstechnologyenergycapitalLabor(output)producedservicesand/orGoodstyproductiviTotal

++++=

Measurement of productivity of knowledge workers It is easy to measure the productivity of quantifiable tasks (e.g., the number of units produced by an individual working on a machine during a given time period). But it is difficult to measure the productivity of tasks (such as pharmaceutical research) that cannot be measured in units. In fact, measuring the productivity of skilled workers is difficult because of its dependence on many intangible and qualitative factors. However, in order to improve planning and control at the organizational level, it is essential to quantify the work and keep track of the productivity of the organization. The productivity of a knowledge worker is much more difficult to measure than that of industrial workers for the following reasons: • The quality of a knowledge worker’s output cannot be determined immediately.

For example, the effects of a strategic decision may not be evident for several years and even then the positive or negative effect of the strategic decision depends on various external factors beyond the control of those responsible for taking the decision.

• Often, the output of knowledge workers contributes only indirectly to the achievement of the end result, and is, therefore, difficult to measure. For example, a human resources manager solves the employees’ problems and attempts to improve the quality of work life of employees. These efforts contribute indirectly to an increase in productivity.

• Knowledge workers often assist other organizational units/departments indirectly and this contribution is usually difficult to measure. For example, a scientist in the R&D department may give a suggestion to the marketing department to highlight a specific attribute of the product while marketing it. In this case, it is difficult to measure the actual contribution of the scientist in the increase in sales of the product.

Measurement of productivity in service organizations Measuring productivity in a service organization is difficult due the intangible nature of the product. Service companies base productivity on the number of tasks performed, or the number of customers served in any given time period. Other measures include a comparison of the service provided with established company, industry, or customer quality standards. Since it is often difficult to establish a standard time for a task in the service industry, usually, only a probable time for a task can be established. For example, the time taken to approve a loan varies from bank to bank. As a result, banks sometimes focus on the speedy approval of loans as a function of their operational efficiency.

In order to measure productivity, service professionals maintain time-sheets to indicate the amount of time spent on a given task. In cases where the task is routine and involves minimal customization, the quantity of work, as for example, the number of service calls made per day, the number of queries handled, or the number of customers served, is used as a measure.


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Productivity and Quality As discussed earlier, productivity and quality are two components of performance. Quality, like productivity, is a key to the well-being of an organization. Quality is complementary to productivity and vice versa. Neglecting one of these can lead to the organization performing poorly. Thus, there should not be any compromise on quality when trying to increase productivity. Just by producing more products or serving more customers, an organization cannot expect to increase profits or retain customers. The product or service should be able to satisfy the requirements of the customers. By improving the quality of a product or service, the organization can improve its competitive position in the market. Many organizations have thus shifted from the production-oriented approach to the marketing oriented one. A focus on quality in the production processes can contribute to improved productivity by reducing wastage in production, reducing the number of rejects, and optimizing the process and procedure. The quality of the product or service can ensure that the organization retains its customers and is simultaneously able to attract new customers. With properly designed operating systems and procedures, an operations manager can improve the quality of the products and the productivity of the factors of production. As opposed to this, an inefficient design leads to a decline in quality and productivity, resulting in overall inefficiency and sub-optimal performance.

The following sections will focus on the concept of quality and its role in organizational success.

STATISTICAL CONCEPTS IN QUALITY CONTROL

(Added under ‘Control Chart for Variables’ in page 268 of the first edition)

Control charts for variables

Control charts for variables are used to maintain quality standards for a process by evaluating measurable product or service variables like thickness, length, tensile strength, and queue waiting time. These charts are used to evaluate the mean and variability of the process distribution. Control charts for variables include X-Chart and R-Chart.

X-Chart X-Chart illustrates the central tendency of the inspected samples. For a meaningful analysis, both the X and R-charts are used simultaneously. The upper and lower control limits for X-chart are given by the following equations:

UCL = x + A2 R

LCL = x – A2 R

Where, x = Mean of means of all the inspected samples

R = Mean of ranges A2 = Constant whose value is dependent on the sample size.

R-Chart or range chart R-Charts show the variability of the process. A process is said to be in control when both the accuracy (mean) and precision are in control. The range chart is used to check the process variability. Range for a sample is given by the difference in the highest and lowest values in the sample. R = xmax - xmin


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Control limits for R-chart is given by the formula:

CL = R

UCL=D3 R

LCL= D4 R D3 and D4 are constants whose values are based on the sample size. Problem 18.1

Table 18.1 gives the means ( x ) and range (R) of ten samples each of size 5.

Table 18.1 Sample 1 2 3 4 5 6 7 8 9 10

x 40 47 38 42 46 42 51 39 41 46

R 6 7 4 8 7 3 6 4 8 6

(Conversion factors for sample size 5 are given as A2 = 0.59, D3 = 2.34, D4 = 0.16)

Comment on the state of the process by using mean and range charts.

Solution The mean of means is given by:

43.210432

nx

x ==∑

=

The mean of range

5.91059

nRR ==∑=

The control limits for the X-chart

UCL = x + A2 R = 43.2 + 0.59 × 5.9 = 46.68

Figure 18.1: R-Chart


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LCL = x - A2 R = 43.2 - 0.59 × 5.9 = 39.7 CL = 43.2 Control limits for the range charts

UCL = D3 × R = 2.34 × 5.9 = 13.8

LCL = D4 × R = 0.16 × 5.9 = 0.944 CL = 5.9

Based on the control limits, R-charts and X-charts are plotted as shown in Figures 18.1 and 18.2 respectively.

Figure 18.2: X-Chart

The process control state is commented on after analyzing both the charts. In range chart all the sample ranges fall within the control limits indicating that the process in control. But the process cannot be assumed to be in control because by observing X-chart we can see the sample means are outside the control limits indicating out of control process. An in control process must satisfy the requirements of both the X-chart and R-chart.

STATISTICAL CONCEPTS IN QUALITY CONTROL

(Added under ‘Control Chart for Variables’ in page 268 of the first edition)

Control Charts for Attributes In control charts for variables, a measurable variable is used like weight, length, width, etc., in the inspection process. But if the quality characteristics are not quantifiable then the items after inspection are identified either as defective or non-defective. In such situations, control charts for attributes are used. Two types of control charts for attributes – the P-chart and the C-chart - are described below.

P-Chart or fraction defective chart p-Chart is employed to find out the proportion of defective items in a selected sample. It is used to control the proportion of defective items being produced by a production process.


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inspected units ofnumber Totalsample ain items defective ofnumber TotalP =

The control limits are given by

UCL= P +3n

)P(1P −

LCL= P - 3n

)P(1P −

Where n= Number of sample inspected, and P = Average of proportion of defective items In case the LCL is negative, it is taken as zero.

Problem 18.2 Based on data given in Table 18.2, construct a p-chart or fraction defective chart. Size of sample is given as 30.

Table 18.2

Sample 1 2 3 4 5 6 7 8 9 10

Number of defectives

3 2 2 1 4 2 1 4 2 1

Solution Fraction defective for each sample is given by

p = c/n = number of defective/sample size

Fraction defective for Sample 1 = 3/30 = 0.1

Fraction defective for Sample 2 = 2/30 = 0.067

Similarly, calculate the fraction defective for all ten samples

Sample Fraction Defective

1 0.1 2 0.07 3 0.07 4 0.03 5 0.13 6 0.07 7 0.03 8 0.13 9 0.07

10 0.03

P =N

P∑ = 0.73/10=0.073

3n

)P(1P − = 330

)073.00.073(1− = 3

30.927)00.073(

= 3 × 0.0475 = 0.142

Substituting these values in the equations for control limits for the p-chart, we get


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CL=0.073

UCL =0.073 + 0.142 = 0.215

LCL =0.073 - 0.142 = -0.069 = 0, as the value obtained for LCL is negative.

The Figure 18.3 of p-chart indicates that the process is in control as all the values are within the control limits.

Figure 18.3: P-Chart

C-Chart or Number of Defects Chart C-Chart is used to illustrate the total number of defect in an item when the item may contain more than one defect. C-Charts are used to find the total number of defects or average number of defects per unit in an inspected sample.

Control limits for a c-chart is given by the equations

UCL = c + 3 c

LCL = c - 3 c Where, c is the average number of defects produced by a process. If LCL value is negative then it is taken as zero, because the number of defects cannot be negative. Problem 18.3 The number of defects in a process was noted after observing 10 samples of a painting job. The errors in these samples are given in Table 18.3.

Table 18.3

Sample Number 1 2 3 4 5 6 7 8 9 10

Number of Defects

3 7 2 0 4 6 1 2 4 3

Using c-chart determine whether the process is in control or out of control.


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Solution

First we have to find the average ( c ), which will be the central limit.

c = nc∑ =

1032

= 3.2

Control limits for this process is given by: UCL = 3.2 + 3 2.3 = 8.57

LCL = 3.2 – 3 2.3 = -2.17 CL = 3.2

As LCL for this sample is negative, it is taken as zero.

Now the C-chart can be obtained by plotting the number of defects on the Y-axis and sample number on the X-axis.

As we see in Figure 18.4 the defects are within the control limits. Hence, the process is treated as in control.

Figure 18.4: C-Chart

CONCEPT OF TQM

(Added under ‘Concept of TQM’ in page 272 of the first edition)

TQM Principles A list of TQM principles is given below that act as a framework for overall improvement in quality standards.

Systematic improvement TQM as a philosophy involves improvement in every aspect of an organization’s functioning, and is not limited to a single product or process. It is a systematic process of improving and maintaining quality standards in an organization. Systematic improvement involves identifying and managing the relationships between various processes and systems to achieve organizational objectives.


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Customer focused The customer is the focus of the total quality management approach, and not an individual product or service. TQM evaluates customers’ requirements, needs and wants and then, provides requisite products and services that satisfy their requirements or exceed their expectations. Customer focus enables an organization to respond effectively, and on time to any changes in markets. TQM also leads to increased customer loyalty, which in turn brings repeat business to an organization.

Continuous improvement Quality is not an end item, but an ongoing process of improvement. Continuous improvement results in improved organizational capabilities, which enable the organization to take advantage of upcoming opportunities. By continuous improvement in the production process, products and employees skills, organizations can maintain their competitive position in the market.

Problem prevention TQM philosophy is proactive rather than reactive. It emphasizes on the prevention of problems rather than the use of remedial measures. This is achieved by employing statistical process control tools, problem solving, system failure analysis, etc. By applying problem prevention principle, an organization can reduce wastages, minimize rejects and improve production process.

Universal responsibility Maintaining quality is not just a prerogative of the top management or quality inspection department but every employee at all levels is equally responsible for its implementation. Quality responsibilities result in the employees being more committed, motivated and creative in performing assigned tasks. Designing quality for products Total quality management principles advocate the integration of quality initiatives with all business processes. Design, the first phase of production, has a significant affect on the overall quality of the end product. Quality in design is important not only for the development of products and services, but also for the development of processes. Genichi Taguchi advocated the importance of quality in design. According to him, activities such as process control, inspection, and defects removal, alone do not improve quality. The three important aspects of design quality are: Meeting customer requirements: During the design process, an organization should focus on meeting customer requirements. As mentioned earlier, making a quality product or service involves providing value to customers, not increasing the number of features. So the design team must either know the exact customer requirements or should be able to estimate them fairly accurately. During the design phase, the design team must be kept informed of any changes in customer requirements, which might influence the design objectives. Capability of process: The product design should take into consideration the capability of the process. During the initial stages of product or service design, operations managers and designers should work together to ensure that production requirements match process capability. Standardization to improve quality: To ensure the quality of products during the design stage, designers should use standard procedures, materials, and processes that have proven quality. Standardization allows easy acceptance of the product in the market. Environment friendly features and attributes in a product or process have become another quality benchmark. For instance, one of the quality benchmarks for automobiles is emission and noise levels.

Facilities and Maintenance Management

• Facilities Management (New) • Necessity of Maintenance Management • Types of Maintenance • Economics of Maintenance • Evaluation of Preventive Maintenance Policies (Modified) • Maintenance Planning (New) • Modern Approaches to Preventive Maintenance • Recent Trends in Maintenance

FACILITIES MANAGEMENT

(New Heading added at the beginning of the chapter) According to Sheila Sheridan, Chairperson (2002-2004), International Facility Management Associationi, in the 1980s, facilities management used to be a practice that involved coordinating the physical facilities, people, and work of the organization using the principles of business administration, architecture, and behavioral and engineering sciences. By the early 2000s, the whole concept had changed and facilities management became a distinct function in management that spanned multiple management disciplines to ensure the smooth and efficient functioning of the workplace and a work environment that integrated people, process, place, and technology. The primary objective of facilities management is to provide a clean and conducive work environment to enable efficient and effective progress of the core functions of an organization, be it manufacturing, distribution, or research. For example, CMC Limited, a leading IT solutions company in India, provides facilities management services to its clients. It undertakes operations and maintenance of IT infrastructure on the clients’ premises. Indian Railways, Indian Oil Corporation Limited, Bank of Baroda, Kodak India, Bombay Stock Exchange, and OTTO India are some of its clients. This section highlights the nature of facilities management, functions of facilities managers, the rationale behind outsourcing of the facilities management function, facilities management and its relationship with maintenance management, and facilities management in India.

Nature of Facilities Management Facilities management has evolved into a distinct function with many organizations across the world creating separate facilities management departments. The facilities management departments carry out the managerial functions of planning, organizing, staffing, coordinating, and controlling with a focus on the activities related to managing the organization’s facilities. The range of activities that is part of the facilities management department is detailed here. Facilities planning and forecasting: Facilities planning, an important function in facilities management is defined as the process of identifying the characteristics of the physical facilities and developing detailed action plans to carry out facilities management activities in the most cost effective way. It includes developing long-term and short-term plans, space forecasting, and financial forecasting.


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Lease administration: This function involves property management. The facilities management department manages all issues regarding leasing out or taking property on lease. Space planning and management: All the activities relating to space allocation, space utilization, space forecasting, and management of space in the facilities are carried out under this function. The space management function ensures that the available space in all the facilities of the organization is utilized properly. Architectural/Engineering planning and design: This function encompasses activities of planning and designing of physical facilities (building, manufacturing plants, showrooms, outlets, etc). Workplace planning, allocation, and management: This function includes workplace planning and design of the workplace based on equipment specification, type of furniture, etc. It also involves estimating the costs involved. This function aims to achieve optimal workplace conditions in order to maximize the work output. Budgeting and Accounting: Financial accounting and budgeting activities (financial forecasting, budget formulation, and execution) are performed under this function. The department looks after the administrative budgets, capital budgets, operations and maintenance budgets, etc. of the various facilities. Real estate management: Activities like site selection, acquisition, building purchase, lease, and disposal are carried out under this function. Construction and project management: Construction activities and project management activities are performed under this function. Construction of facilities like plants, buildings, offices, (storage tanks in case of petroleum and chemical companies) are some examples. Alteration, renovation and workplace installation: Existing facilities have to be upgraded or modified with the changing business environment and strategies of the organizations. These alteration and renovation works are managed by the facilities management department. Communication management: Installation of telecommunication networks, their operations, and maintenance are carried out under this function. Installation of LAN, maintenance of communication systems like EPABX, Internet connectivity, etc., are some examples in this function. Operations, maintenance and repair: Activities like preventive maintenance and remedial maintenance fall under facilities maintenance. Other activities under this function include trash removal, hazardous waste removal and management, energy management, inventory management, repairs, disaster recovery planning & management, and purchase management. Usually, these functions are outsourced to external agencies to reduce the cost of operations. But some organizations retain the discretion to perform certain functions in-house to avoid the involvement of external agencies.

Functions of Facilities Managers Facilities managers perform a wide array of functions. Their basic function, like that of managers of other functional departments, is to develop and execute strategies that fall in line with the organizational goals and objectives and help in achieving them. The primary function of a facilities manager is to organize the necessary services for all facilities of the organization at reasonable costs. Services to provide a conducive work environment through tidy and hygienic surroundings include: housekeeping, maintenance services, plumbing, communication networks, inventory management, and waste management, which have been mentioned earlier (Exhibit 19.1 describes how waste management has become an important function in healthcare institutions). Facilities managers are expected to be flexible as they have to work overtime or at odd


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hours to fix problems that arise in the facilities. For example, leaks in the pipelines or a short circuit leading to fire accidents may disrupt work in the organization. Facilities managers have to see that such problems are corrected as and when they are detected. The following are some of the important functions of a facilities manager. • Conducting and participating in planning activities along with architects,

production engineers, and the top management while performing alterations to facilities or expanding the facilities. This helps in designing better facilities.

• Reviewing the techniques that are used to reduce costs and raising maintenance levels in the facilities. Reduced costs result in improved profits. Better maintenance levels help increase the life of assets and ensure a productive work environment.

• Recruiting and/or hiring people for carrying out various facilities management functions. Manning the department with the right kind of people is an important function of the facilities manager, because the right people in the right jobs help to increase efficiency, and lower wastage and costs.

• Measuring the work performance of the facilities management teams using work measurement techniques and motivating them to raise their performance levels. A highly motivated team always produces better performance and higher productivity

• Developing and maintaining a controlling center that acts as a strategic support to the facilities management department. Such a center acts as the center-point of activity from where the facilities managers can execute and control decisions in the facilities management department.

• Communicating with other department heads and personnel and understanding their nature of work. Also, maintaining good rapport with them as the success of the facilities management department’s functions is dependent on other departments.

• Developing good relationships with the service providers and maintaining a congenial work atmosphere in the facilities by applying sound management principles.

• Guiding employees and the top management in developing innovative ways of improving the work atmosphere in the facilities.

• Acting as change managers in situations where existing processes or infrastructure is modified or replaced with new ones to suit the organizational goals. Facilities managers should make the workers understand the necessity of the changes brought about in the organization and make them work toward achieving the organizational goals.

Exhibit 19.1 Waste Management in Healthcare Industry

Waste management in healthcare institutions is an important and critical activity. Most of the healthcare institutions in India do not have waste management centers nor do they maintain standards in managing the waste in their facilities. Waste management is one of the functions of facilities management. It includes removal of waste from the premises, segregating the waste, disposing it of, and/or treating it to make it harmless to society. Waste in healthcare institutions like hospitals, clinics, etc., include used dressings, disposable syringes, outdated drugs, empty drug bottles, etc. This function helps keep the hospital premises neat and tidy to ensure a cleaner environment and the speedy recovery of patients.


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Bangalore-based Health Care Waste Management Cell (HCWMC) is in the process of educating healthcare institutions to set up in-house waste management centers. It is necessary to have a waste management center in every healthcare institution to manage the waste effectively. According to the HCWMC, healthcare institutions lack the required information and support to manage healthcare wastes. Realizing this, the HCWMC has set up a resource center in Bangalore to train healthcare institutions in waste management, capacity building, and providing access to required information. The mission of the HCWMC is to advocate the cause of managing medical waste, disseminating information, training, research, capacity building, and networking. The training programs and supporting information aim at encouraging and helping the institutions to set up their in-house waste management centers. The HCWMC feels that it is necessary to bring about an attitudinal change toward waste management in the people managing these institutions and it has worked to bring in institutional commitment and participation from all levels of people in these institutions.

Further, stringent rules from the government regarding waste management and growing demand for quality from insurance companies (for accreditations) have also forced the healthcare institutions to take up in-house waste management in their facilities.

Adapted from Vijaya, K. “In-House Systems for Waste Segregation is Critical: Interview with Dr.D. Gopinath.” Express Healthcare Management. January 15, 2002, October 3, 2006.

Outsourcing the Facilities Management Function Facilities managers use out-tasking/outsourcing to carry out the various tasks in facilities management. Out-tasking (i.e. hiring of individual, specialized vendors) is used more frequently than outsourcing (i.e. hiring of full-service, single source vendors). Out-tasking supplements in-house delivery capability with sub-contracted expertise or capacity. Outsourcing seeks to integrate the expertise and capacity of a service provider toward a strategic objective. Outsourcing the facilities management function has become a part of the strategic planning of organizations to gain long-term benefits. There are many reasons as to why facilities management services are outsourced, one major one being to concentrate on the core competencies and leaving the additional functions to external sources specializing in this field. For example, Punj Lloyd, an engineering construction and facilities management company, provides operational and maintenance services for the power plants of the Oil and Natural Gas Commission (ONGC). By outsourcing these services to Punj Lloyd, ONGC can focus on its core activities of oil exploration and extraction. Another reason why organizations outsource facilities management services is that they don’t have employees with requisite skills. Outsourcing is also done when the organization feels that it is costly to handle the services in-house.

Services that are most commonly outsourced are architectural design, trash and waste removal from the premises, housekeeping, landscape maintenance, property appraisals, hazardous material removal, furniture movement, and food and refreshment services. Johnson Controls is a global operator in this domain and provides a comprehensive list of such services. They include services ranging from budget management and risk management, performance management, operations and maintenance services, product and facility design services. (Refer to Exhibit 19.2 for further details on Johnson Controls).

Generally, the need to outsource facilities management originates from the facilities management department itself. Facilities managers do this to reduce their cost of operations and improve the quality of tasks performed. The benefits and costs accrued are not accepted by everyone. Some sections accept outsourcing as a means to reduce costs while some believe that it increases costs. Let us look at the costs and benefits associated with outsourcing of services in facilities management.


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Exhibit 19.2 Johnson Controls – One Stop Shop for All Facilities

Management Needs Johnson Controls is a global player in the field of electrical and electronic control systems like thermostats, actuators, sensors, valves, etc. Facilities management is one of the services that the company offers. It provides various services of facilities management at the strategic level as well as at the operational level, under one roof. Strategic level services include Account/Budget Management, Supply Chain/Supplier Diversity, Performance Management, Triple Bottom Line Reporting, Risk Management, Transition Management, Strategy Development, Subcontractor Management, Capital Planning, and Benchmarking. Operational level services are classified into three broad categories – business site services, business critical and technical services, and workplace and infrastructural design services. Several services are performed under each of these. For example, reception, conference services, surveillance, reprographics (duplication of printed material), and waste management services are under business site services. Johnson Controls has developed support mechanisms to ensure quality services to clients enhance efficiency, and reduce costs. They include: Center for Excellence: As part of offering account/budget management services, the company has established a global center for excellence where experts from different fields analyze the client’s existing accounts and industry benchmarks to reduce operational costs. Training Programs: Johnson Controls provides training to its employees as well as those of clients in the areas of leadership, ethics, safety, and technical skills. Customer Reports: In addition to the various services offered, it also provides reports on daily, monthly, and quarterly basis to all levels in the client’s organization. This is to ensure that the client can see the return on investment gained in outsourcing various services to the company. The reports essentially contain Johnson Controls’ performance on the client’s premises that enable the client to make effective business decisions. Clients for Life: It is a customer relationship management program which focuses on account management and creativity workshops so that clients’ needs and requirements are identified and understood in the face of the ever changing work environment.

Adapted from www.johnsoncontrols.com. Costs associated with outsourcing Outsourcing of facilities management to external agencies is not always as profitable as the facilities managers and the top management expect. There are also different associated costs, which can hinder the expected savings from the outsourcing contracts and reduce the quality of performance. Some of these costs are listed here:

Loss of Control: If the client organization passes over control of its facilities to an external agency, it is like inviting the agency to gain monopoly over its facilities. This can lead to loss of control over crucial decisions like alterations or expansion of facilities or cost savings. Also, the client may lose the power to control any deviations in the performance of the vendors. Loss of control can lead to strained relationships between the client and vendors and become the cause for serious conflicts. To avoid such situations, client organizations generally enter into a contract with multiple vendors to restrict the monopoly situation and to have better control over the vendors’ performance.

Decrease in flexibility: Switching costs arise when a client enters into a contract with the vendor(s) and hands over control. For instance, when the facility management of a manufacturing plant is outsourced, the client organization first decommissions all the machinery and then the vendor takes over. As a result, the initial costs are very high


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for the service provider. To make up these costs and gain profits in the operations, the service providers need some time. Thus, they would normally prefer long-term contracts with the client organizations (as long-term contracts would be profitable to the vendors). But such long-term contracts are usually less flexible for the client organizations, especially if they want to regain control. Also, the cost to the client will increase in a long-term contract as the vendors have the flexibility to raise or revise prices.

Decrease in staffing quality: The vendors bring in their own staff and workers who are trained to work in multiple work places (to perform the facilities management activities in different client organizations). Sometimes vendors employ skilled workers in the initial period of the contract. Once the client is satisfied with the performance of the vendor, these skilled personnel will be replaced with semi-skilled or unskilled personnel to reduce the direct operating costs. This can lead to an increase in costs for the client and a decrease in the quality of work.

Increase in costs over time: The goals and objectives of the client and service provider differ widely and may not be in line with each other. The service providers’ need to raise profits can lead to exploitation of the client’s resources or a reduction in the quality of work. They may also exclude some of the processes from the contract with the excuse of raising costs and charging for them separately. This will increase the costs of the client and eventually decrease the quality of work.

Benefits of outsourcing Despite these costs associated with outsourcing the facilities management function, there are also certain advantages or benefits that accrue to the client organizations. The increasing popularity of outsourcing is evidence of the associated benefits. Hence, many organizations (large and small) have outsourced the management of their facilities to specialized vendors in this field. Given here are some of the benefits that clients derive by outsourcing the management of their facilities. Reduced costs: When the management of facilities is outsourced to external vendor(s), the cost of operations go down considerably as the organization (client) need not pay regular salaries for the employees and the wages of the workers involved in these activities. The client has to just pay the pre-agreed price for the services offered by the service provider. Generally this is lower than the sum of the actual costs of operations. Besides, the client need not spend money on training the workers and that also leads to cost savings. Increased quality: Services are usually outsourced to specialized service providers. They bring in staff and workers who are trained to carry out the tasks. Hence, the quality of work performed by these trained workers is usually far superior to that of the client.

Focus on core competencies: This is one of the major reasons for the increasing popularity of outsourcing. The basic goal of any organization is to satisfy their respective customers. By outsourcing the functions of facilities management, organizations can focus more attention on their core competencies, and this leads to better customer service. It also increases their competitive advantage in the respective markets.

Facilities Management in India The scope of facilities management in India has changed a lot in the period between 1980 and 2000. Earlier, facilities management meant only keeping the premises clean and most organizations, whether public or private sector, had their own departments to


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look after this function. This changed with the entry of multinational companies in the 1990s after the liberalization of the economy, especially with the entry of multinational software companies, which had their own views on cleanliness levels. The next change occurred with the entry of multinational facilities management companies like C B Richard Ellisii, Hadeniii, Knight Frankiv, and Sodexov.

Higher levels of cleanliness have become the norm now and many facilities management service providers specializing in various services have entered the market. The market for outsourcing facilities management services has also grown. Apart from janitorial and maintenance services, other complex services like waste management, water treatment and management, asset protection services, health facilities management, etc. have come under the scope of facilities management. Updater Servicesvi (UDS) and Integrated Property Management & Servicesvii (IPMS) are some examples of Indian companies that provide such services.

Facilities management services are required by different clients both in the manufacturing sector as well as in the services sector. Facilities management services in the manufacturing and services sectors vary in terms of scope. The presence of heavy machinery and equipment in manufacturing companies’ calls for preventive maintenance services and safety & security services, while service oriented companies like banks and BPOs may see janitorial services as the prime service under facilities management. But many services are common like waste removal, food and catering, landscape maintenance, janitorial services, security services, etc. As a result, many companies provide services to both manufacturing and service organizations. For example, UDS, specializes in providing janitorial services to manufacturing as well as service organizations like Godrej, Reliance, Infosys, American Express Bank and Saint-Gobain.

Facilities Management and Maintenance The facilities maintenance is a component of facilities management. When facilities maintenance is done by in-house staff, it is the duty of the facilities manager to see that the facilities are properly maintained. In large organizations, facilities managers recruit dedicated maintenance staff or bring in outsourced staff to perform maintenance tasks throughout the year at different locations. In smaller organizations, the regular staff is assigned the additional work of maintenance as there is not enough maintenance work to keep a dedicated team busy throughout the year. People with specific talents in carpentry, plumbing, gardening, troubleshooting of equipment, etc., are recruited to be part of the maintenance teams in various organizations. Maintenance of facilities can also be outsourced as part of the facilities management strategy.

The functions of facilities managers under the maintenance function include carrying out periodical preventive maintenance checks on machinery, checking for safety in the work environment, and allocating budgets for upgrading the environment of the facilities. Preventive maintenance (discussed in the later part of the chapter), is outsourced by many organizations to service providers who are specialists in this field. A facilities manager should have proper knowledge of all the facilities in the organization and the machinery and equipment that are installed in various departments.


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EVALUATION OF PREVENTIVE MAINTENANCE POLICIES

Added under ‘Evaluation of Preventive Maintenance Policies’ in page 283 of the first edition) Problem 1

A workshop has 50 identical machines. The failure pattern of these machines has been observed as follows over a year.

Elapsed time after last maintenance in months

1 2 3 4 5 6 7 8 9 10 11 12

Probability of failure

0.05 0.05 0.10 0.10 0.10 0.10 0.10 0.10 0.10 0.10 0.05 0.05

If the yearly cost of servicing the failed machines is Rs.7000, what is the average cost of repairs per machine per occasion?

Solution Mean time between failure = ∑ i.P(i) = (1 x 0.05) + (2 x 0.05) + (3 x 0.10) + (4 x 0.10) + (5 x 0.10) + (6 x 0.10) + (7 x 0.10) + (8 x 0.10) + (9 x 0.10) + (10 x 0.10) + (11 x 0.05) + (12 x 0.05) = 6.5 months Number of breakdowns per year = 12/6.5 = 1.846 Number of breakdowns per year for 50 machines = 1.846 x 50 = 92.3 Average cost of repairs per machine per occasion = 7000/92.3 = Rs.75.84 Problem 2

The probabilities of failure after maintenance for a machine are given in the following table

Quarters after maintenance 1 2 3 4 5 Probabilities of break-down 0.1 0.1 0.2 0.3 0.3

If there are 30 identical machines in the plant, what is the expected number of breakdowns in the 3rd quarter? If the average cost of preventive maintenance is Rs.500 and the cost of remedial maintenance per machine is 4000, calculate the average total cost of maintenance per month.

Solution

Expected number of breakdowns Bt = N (P1 + P2 + P3 +…+ Pt) + Bt-1 P1+ Bt-2 P2+…+ B1 Pt-1

Given N=30 and t=3

B1 = N P1 = 30 x 0.1 = 3

B2 = N(P1 + P2) + B1P1

= 30(0.1 +0.1) + 3 x.0.1

= 6 + 0.3 = 6.3

B3 = N(P1 + P2 + P3) + B2P1 + B1P2

= 30(0.1 + 0.1 + 0.2) + 6.3 x 0.1 + 6 x 0.1


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= 12 + 0.63 + 0.6

= 13.23

Hence, the expected number of breakdowns in the 3rd quarter = 13.23

Total cost of maintenance for every 3 quarters = CP x N + CR x Bt

= 500 x 30 + 4000 x 13.23

= 15000 + 52920

= Rs.67920

Average total cost per quarter = 67,920/3 = Rs.22,640

Average total cost per month = 22,640/3 = Rs.7546.67

MAINTENANCE PLANNING

New Heading added after ‘Evaluation of Preventive Maintenance Policies’ in page 283 of the first edition) As the production costs are increasing and the manufacturing processes are becoming more complex, the need for maintenance has increased manifold. Planned maintenance helps organizations reduce the expenditure on maintenance and the costs associated with failures. Preventive maintenance activities are repetitive in nature and performed according to a schedule. On the other hand, remedial maintenance is not repetitive and is performed on a need-to-do basis, as future breakdowns cannot be predicted. Preventive maintenance activities involve identification of probable problems and estimation of the amount and type of work that is required to keep the systems working in optimal working condition. This estimation enables maintenance managers to schedule and plan their maintenance programs. Subsequently, they determine the sequence in which various maintenance activities are carried out. Managers use the available information to determine the labor hours, tools, equipment, materials and other accessories required to implement the maintenance plan. The maintenance tasks are prioritized on the basis of their significance in maintaining a continuous production process when maintenance resources are scarce. These priorities help organizations allocate resources to those critical systems whose failure will have disastrous effect on the production process. Besides, maintenance managers often maintain inventories of equipment and materials required for maintenance activities to reduce the lead-time before a repair can begin. Organizations with production facilities in more than one location reduce inventory costs by storing basic and frequently used maintenance materials at each location, and occasionally used materials at a centralized location. Many organizations use computerized maintenance management systems to improve their maintenance planning. Maintenance is carried out throughout the life of a piece of equipment. The bathtub curve assists in decisions regarding maintenance requirements. If organizations are not able to fulfill the maintenance requirements, they outsource maintenance activities from outside vendors.

Bathtub Curve

It would be of great help to operations mangers if they were able to predict when the probability of equipment failure will be highest. The bathtub curve illustrates the different stages a piece of equipment goes through in its lifetime and the probability of failure during a particular stage. Figure 19.2 illustrates the structure of a bathtub


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curve. During the first stage of infant mortality, the probability of failure is very high but it decreases rapidly. The reason for these failures can be attributed to improper design and installation. This stage is also known as the ‘burn in’ stage where the functional defects in the products come to fore. Once this stage is passed, the equipment enters into the adult or useful life stage where the failure rate is constant, and to some extent, predictable. Proper maintenance of equipment can ensure the longevity of this stage. Most of the causes of failure during this stage are attributed to external causes or accidents, for example, a mistake by an operator, or usage of improper materials. Then finally, the equipment reaches old age or wear out stage, where the failure rate increases. During this stage, organizations go in for a complete overhaul of the system or replacement of the system. The failure rate in each of these stages depends on the reliability of the equipment installed and on the amount of maintenance effort spent on it. The bathtub curve can be used not only to study the failure rates characteristic of new equipment but can also be applied for old equipment, which has undergone servicing under preventive maintenance. After the preventive maintenance is completed, the equipment again goes through the stages defined in the bathtub curve. The bathtub curve may not always be an effective indicator of the failure rates of equipment. For example, present day manufacturers dispatch equipment only after the ‘burn in’ period is over, so that they can identify and rectify the problem before it reaches the end users. So the initial stage of the bathtub curve will not be applicable to such machines.

Figure 19.2: Bathtub Curve

Contract Maintenance

In order to reduce the costs incurred in maintenance, many organizations outsource their maintenance activities. The process of outsourcing an organization's maintenance activities to external vendors is known as contract maintenance. Contract maintenance is useful for seasonal or irregular maintenance activities, which do not warrant the involvement of the in-house maintenance department. Further, external maintenance personnel have the requisite training and expertise to service complex systems. By outsourcing non-core activities like maintenance, an organization can concentrate on its core business activities.


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i International Facility Management Association (IFMA) is a professional association for

facility management with more than 18,500 members. The members represent 125 chapters in 60 countries. IFMA conducts research, provides educational programs, certifies facilities managers and conducts international conferences on facilities management.

ii CB Richard Ellis is a global full-service real estate company. It has operations in 58 countries with headquarters in California, USA.

iii Haden, based in England, was established in 1816 as a heating and ventilating company. Later it began to provide facilities management services. It is a specialty provider in building management services.

iv Knight Frank is an international real estate and property consultant having operations in over 30 countries with 165 offices.

v Headquartered in France and with operations in 76 countries, Sodexo is an international facilities management services operator. Its services include Food and management services, Service vouchers and cards, River and harbor cruises.

vi Updater Services is a Chennai-based integrated facilities management service provider. It specializes in building maintenance services, M&E services, HVAC maintenance services, and garden & lawn maintenance services.

vii Integrated Property Management & Services is a subsidiary of Infrastructure Leasing and Financial Services Limited (IL&FS), which focuses on facilities management.

Project Management • Necessity of Project Management

• Network Modeling

• Project Planning Methods (Modified)

• Project Crashing

PROJECT PLANNING METHODS

Critical Path Method (CPM) Computation of latest times: backward pass (Added under ‘Computation of latest times: backward pass’ in page 294 of the first edition) Problem The time estimates for completing the construction of a facility project are shown in the figure below. Determine the critical path of the project.

Figure 20.4: Network Diagram for Problem 20.2

Solution The critical path can be computed by knowing the earliest and latest times for each of the activities mentioned in Figure 20.4. The following table shows the earliest and latest times.

The earliest start time (EST) for event 1 in the network will be 0 or E1 = 0, E2 = E1+D1-2, E3 = E1+D1-3, etc.

Earliest finish time (EFT) = Activity duration + EST

Table 20.3: Calculation of Earliest and Latest Time for an Activity Activities Duration

(Days) (D)

Earliest Starting Time (E)

Earliest Finish Time (E+D)

Latest Finish Time (L)

Latest Starting Time

(L-D)

Float (L-D)-E

1-2 6 0 6 6 0 0 1-4 4 0 4 6 2 2 2-3 3 6 9 9 6 0


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3-5 5 9 14 12 7 -2 3-6 3 9 12 8 5 -3 3-7 6 9 15 15 9 0 4-6 6 4 8 8 2 -2 5-7 3 14 17 15 12 9 6-7 7 10 17 15 8 -2 7-8 3 15 18 18 15 0

In this problem, the float value is zero for activity 1-2, 2-3, 3-7, and 7-8. Hence, the critical path for this network problem is 1-2-3-7-8.

PROJECT PLANNING METHODS

PERT (Added after Problem 2 under PERT in page 298 of the first edition)

Problem For the network diagram given in Figure 20.4, the optimistic, pessimistic, and most likely times are given as under. What is the probability of the project taking more than 20 weeks for completion?

Activities Optimistic time to

Pessimistic time tp

Most likely time tm

1-2 4 8 6 1-4 4 6 5 2-3 2 5 4 3-5 5 7 6 3-6 3 6 4 3-7 5 9 6 4-6 4 9 6 5-7 2 5 4

6-7 6 9 7 7-8 3 6 5

Solution To find out the probability, one has to find the expected mean time Te and standard deviation.

Activities Optimistic time to

Pessimistic time tp

Most likely time tm

Expected mean time te

( )6

4ttt mpo ++

Variance 2

op

6tt

−

1-2 4 8 6 6 0.4444

1-4 4 6 5 5 0.1089 2-3 2 5 4 3.83 0.25

3-5 5 7 6 6 0.1089

Project Management

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3-6 3 6 4 4.17 0.25

3-7 5 9 6 6.33 0.4356

4-6 4 9 6 6.17 0.6889

5-7 2 5 4 3.83 0.25

6-7 6 9 7 7.17 0.25 7-8 3 6 5 4.83 0.25

Completion time TE = sum of the all the (expected mean times (te) on the critical path as determined in problem 2 (1-2-3-7-8) = 6 + 3.83 + 6.33 + 4.83 = 20.99 (21 weeks approx)

Standard deviation (of critical path) = √variance = √ 0.4444 + 0.25 + 0.4356 + 0.25 = √1.1747

Probability of project extending over 20 weeks = (20 – 20.99) / (√1.1747) = 0.99/1.17 = 0.846

Thus, there is a very high probability of the project going on beyond 20 weeks.

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